convolution discrete and continuous time-difference equaion and system properties (1)

Discrete Time Signals

Convolution of Discrete Time Signals

Properties of the Systems

B.S. Panwar

Convolve: It is latin word which means fold over or twisting together

Sampling Property

Continuous Time

&

Discrete Time

Properties of Unit Impulse Function

f ( t ) ( t ) = f ( 0 ) ( t )

f ( t ) ( t - T ) = f ( T ) ( t - T)

Multiplication Properties

In this case the amplitude of (t) may be 1 / 2 and its duration may be 2

secs. Making the area to be unity. In this case you will have the pulse of

amplitude { f(0) / 2 } and its duration of 2 sec. Since, the pulse is of

finite duration, so its response will be affected by the pulse response.

Sampling Properties of Unit Impulse

Sampling Properties

=∫- ∞

f ( t ) (t) dt

∞=∫

- ∞

(t) dt

∞

f ( 0 ) f ( 0 )

=∫- ∞

f ( t ) ( t - T) dt

∞=∫

- ∞

( t – T ) dt

∞

f ( T ) f ( T )

When you do sampling, we are going to get the sample value at

unique time t = T and it will have the sample value f (T).

Sampling Property of Discrete Impulse

[ n ] =

1 For n = 0

0 For n ≠ 0

[ n - no] =

1 For n = no

0 For n ≠ no

x [ n ] [ n ] = x [ 0 ] [ n ] Sample magnitude and location

x [ n ] [ n - no ] = x [ no ] [ n - no]

Sampling Property

Interpretation of u[n] and u[ n – k ]

n

u [ n ]

0 1 2

k

u [ - k ]

0-4 -2

The function u [n ] = 1 for n > 0

= 0 for n < 1

The function u [n – k ] = 1 for [n – k ] 0 or n k

= 0 for [n – k ] < 0 or n < k

u [ - k ] = 1 for (k-n) ≤ 0 or n k now put n= 0

we get 0 k or k 0

= 0 for k > 0

Let us consider the function for n = 0. The

function will be u [-k]. Now this function

has independent variable as “k”

This is signal u [n]

delayed by k samples

Here independent

parameter is n and k is

the search parameter

Interpretation of u[n-k] and u[ – k +n ]

Here independent parameter is “n” and “k” is the search parameter

So the dependence of discrete time signal x[n[ is seen on “n” the independent parameter,

which is the first term in the brackets and “k” is the search parameter. Here u [ n – k ] is

seen the signal delayed by “k” samples as shown below. This corresponds to delay of “k “

samples of u [n]. Here u [n – k ] signal is viewed as “k” sampled delayed u[n] signal.

n

u [ n – k ]

0 1 2

The function u [n ] = 1 for n k

= 0 for n < k

k

n

u [ n ]

0 1 2

The function u [n ] = 0 for n < 0

= 1 for n > 1

Interpretation of u[n-k]

Interpretation of u[n-k] and u[ – k +n ]

k

u [ - k ]

0-4 -2

u [ - k ] = 1 for or k 0

= 0 for k > 0

Here “k” is the is the

independent variable.

Here independent parameter is “k” and “n” is the search parameter

So the dependence of discrete time signal x[k[ is seen on “k” the independent parameter,

which is the first term in the brackets and “n” is the search parameter. Here u [ - k + n ]

is seen the signal advanced by “n” samples as shown below. So we first visualize first u [-

k] and than visualize u [ - k + n]

Interpretation of u[- k + n]

k

u [ - k + n ]

-2 0

The function

u [-k + n ] = 1 for k n

= 0 for k < n n- 4

Interpretation of u[n] and u[ n – k ]

u [ 1 - k ] = 1 for k 1 ( ( 1 – k ) 0 or 1 k )

= 0 for k > 1

Let us consider the function for n = 1.

The function will be u [ 1- k]

k

u [ 1 - k ]

0-4 -2 1

k

u [ n - k ]

0-4 -2 1

For n > 0. That is n is a positive real number

u [ n - k ] = 1 for k < n { ( n – k ) 0 or

n k or k n}

= 0 for k > n

k0-4 -2 1

For n > 0

u [ n - k ] = 1 for k n

For n < 0For n < 0. That is n is a negative real number

u [ n - k ] = 1 for k n { ( n – k ) 0 or

n k or k n}

u [n-k] = 1 for k n

Discrete Time Signal Linear Combination

of Weighted Shifted Unit Impulse

x [ n ] [ n +1] = x [ -1 ] [ n +1 ]

x [ n ] [ n ] = x [ 0 ] [ n ]

x [ n ] [ n - 1 ] = x [ 1 ] [ n - 1 ]

x [ n ] [ n -2 ] = x [ 2 ] [ n -2 ]

- k

]k -n [ ]k [ x [n]x

A discrete signal is linear combination of weighted shifted impulses

Interpretation of Discrete Time Convolution

y [ n ] = x [ n ] [ n ]x [ n ]

]n[x]k -n [ δ [k]x [n]y n

-k

That means if we input [n] to the system being characterized by x [n] . Than w get

x [ n ] [ n +1] = x [ -1 ] [ n +1 ] = y[-1]

x [ n ] [ n ] = x [ 0 ] [ n ] = y[0]

x [ n ] [ n - 1 ] = x [ 1 ] [ n - 1 ] = y [1]

x [ n ] [ n -2 ] = x [ 2 ] [ n -2 ] = y[2]

This uniquely characterizes the system response which is x[n]. This

is the response to an impulse. That means the response [n]

uniquely characterizes the system.

Specific Case when x [n] = 1 for n 0

]n[u]k -n [ δ [n]y 0k

y [ n ] = [ n ] * x [n] = u [n]

y [ n ] = x [ n ] [ n ]x [ n ]

That means, if I want to know about the system. I start an impulse from [ n + k ], That

means k starts from - and varies up to infinity. In this process I can now about the system.

Let us assume x [ n ] = u [n]. In this case x [k] = 1 for k 0. So the expression we will get is:

]n[x]k -n [ δ [k]x [n]y n

-k

If x [ n] =1 for n 0 than

That means if we input [n] to

y [ n ] [ n ]h [ n ]

]n[h]k -n [ δ [k]h [n]y n

-k


y [ n ]x [ n ]h [ n ]

n

-k

k n] [ h [k]x [n]y

This implies that output of the system for an input x [ n ] will be superposition of

the scaled version of delayed shifted impulses [ n – k ]. Let the response for [n-

k] is hk [ n ]


n

-k

k n] [ h [k]x [n]y

Let h0[n] is the response to [n]

0 1

1 ho [ n ]

n2

Than h-1[n] is the response to [n+1]which is the

response for unit advanced impulse. So the response

h-1[n] will be unit advanced response as shown1

1 h-1 [ n ]

n0 1

0 2

1h2 [ n ]

n3 4

1

Let h1[n] is the response to [n-1]which is the

response for unit delayed impulse. So the response

h1[n] will be unit delayed response of ho[n] as shown

3

0 1

1

h1 [ n ]

n2

Identically h2[n] is the response to [n-2],which is the

response for two unit delayed impulse. So the response

h2[n] will be two unit delayed response of ho[n] as shown

h-1[n], h0[n], h1[n], h2[n]

Is the response to [n = 1], [n], [n-1]

and identified as hk[n] than y [n] can be

written as:


n

-k

k n] [ h [k]x [n]y

In general h-1[n], ho[n], h1[n] and so on are the responses for [n +1], [n ]

[n -1]. That means the most general form defining hk[n] is the responses

for an input [n-k]. These responses may not be same as the time shifted

version of h0[n] (if the system is not LTI), for the leading or delaying

behaviours. But the system is considered to be time invariant. Therefore,

hk[n] are the time shifted version of h [n] which are obtained for the input

[ n – k ]. Therefore, these will be h [ n – k] as shown in the figure of earlier

slide.

Concept of Discrete Time convolution

n0 1

x [ n ] 0.5

2.0

n0 1 2

h [ n ]

- k

]k -n [h ]k [ x [n]y

y[n] = x[0] h[n] + x[1] h [n-1]

y[n] = 0.5 h[n] + 2 h [n-1]

When the input x[n] is applied we get the output x[0] h[n].

Than we get the input x[1]. This is the way sequence is going

to flow in. This is implied that we will get the response h[n]

delayed by unit pulse. In this case we have taken the input x[-

k] This will e multiplied by h [n-1]. That is x[1] h [n-1].

y[n] = x[0] h[n] + x[1] h [n-1]

Physically it means that input will be x[o] than x[1]

First sample no. at n=0 Last sample no at n=3

Duration of convolved output = (2+3) – 1 = 4

1.0

n0 1 2

y[ n ]

0.5

2.52.5

2..0

- k

]k -n [h ]k [ x [n]y

Let us find the response for x [ n ] which has only two samples

Discrete Time Convolution: Concept

0 1

0. 5

2x [ k ]

k

1]-n [h ] 1 [ x [n]h [0] x ]k -n [h ]k [ x [n]y 1

0 k

= 0.5 h [n] + 2 h [n-1]

0 1

1h [ k ]

k2

x [ n ]

0 1

1h [ k ]

k2

2 multiplied

0.5 multiplying to h [n]

+

+ 2 multiplying to h [n] after 1 unit delay that is h [n-1]

0.5 multiplied

1 2

1 h [ k -1 ]

k30

1 2

2.5 y [ n ]

k3

0.52.0

So the output starts at 0 and goes up to sample no. 3

Total no. of Sample in Convolution Output

n1 2

x [ n ] 0.5

2.0

k0 1 2

h [ k ]

3

n0-2

x [-k ] 0.5

2.0

-1

So the no. of samples in the convolved output will be

: 5 – 2 + 1= 4

The last sample value in the convolved output is n = xh + h hHere xh and h h are the values in the un-inverted pattern

So the no. of samples are: ( xh – xl) + ( h h - hl) + 1

: (N1 – 1)+ (N2 – 1) + 1 = N1 + N2 -1

So the convolved output extends from sample no.:

= ( xh + hh ) - ( x l + h l )

So the no. of samples are: ( xh + hh ) - ( x l + h l ) + 1

The first sample value in the convolved output is n = xl + h lHere xl and h l are the values in the un-inverted pattern

Discrete Time Convolution

First and Last Sample Value

in the Convolved Output

Total no. of Sample in Convolution Output

n1 2

x [ n ] 0.5

2.0

k0 1 2

h [ k ]

3

n0-2

x [-k ] 0.5

2.0

-1

So the no. of samples in the convolved output will be

: 5 – 2 + 1= 4

The last sample value in the convolved output is n = xh + h hHere xh and h h are the values in the un-inverted pattern

So the no. of samples are: ( xh – xl) + ( h h - hl) + 1

: (N1 – 1)+ (N2 – 1) + 1 = N1 + N2 -1

So the convolved output extends from:

= ( xh + hh ) - ( x l + h l )

So the no. of samples are: ( xh + hh ) - ( x l + h l ) + 1

The first sample value in the convolved output is n = xl + h lHere xl and h l are the values in the un-inverted pattern

Concept of Discrete Time convolution

n0 1

x [ n ] 0.5

2.0

n0 1 2

h [ n ]

- k

]k -n [h ]k [ x [n]y

y[n] = x[0] h[n] + x[1] h [n-1]

y[n] = 0.5 h[n] + 2 h[n-1]

When the input x[n] is applied we get the output x[0] h[n].

Than we get the input x[1]. This is the way sequence is going

to flow in. This is implied that we will get the response h[n]

delayed by unit pulse. In this case we have taken the input x[-

k] This will e multiplied by h [n-1]. That is x[1] h [n-1].

y[n] = x[0] h[n] + x[1] h [n-1]

Physically it means that input will be x[o] than x[1]

First sample no. at n=0 Last sample no at n=3

Duration of convolved output = (2+3) – 1 = 4

1.0

n0 1 2

y[ n ]

0.5

2.52.5

2..0

- k

]k -n [h ]k [ x [n]y

Discrete Time Convolution: Concept: Alternate way of looking at it

- k

]k -n [h ]k [ x [n]y

The sum can be carried out on the number of samples in x [ n ]

0 1

0. 5

2x [ k ]

k

Now we can find the convolution for x [k]. That means

we take the contribution of 2 samples. x [0] and x[1]

1]-n [h ] 1 [ x [n]h [0] x ]k -n [h ]k [ x [n]y 1

0 k

This implies that we multiply h [n] containing 3

samples by x [0]. This gives h[0], h[1], h[2] all 0.5.

Now we multiply h[n-1] one sample shifted

version of h[n] by x[1]. This gives h[-1], h[-2]

h [-3] all multiplied by 2. So the response extends

from n = 0 to n = 3, which is ( N1 + N2 -1)

= 0.5 h [n] + 2 h [n-1]

0 1

1

h [ k ]

k

2

0 1

2 3

x[n]

n

1

2

1

-1 0 1

h[n]

n

22

0 1 2

3

y[n]

n-1

4 5 6

2

4

2 2

-2

Convolution of Two Discrete Time Signals

The lower starting sample will be sum of

lowest sample points of two signals

- k k]-n [h [k]x [n]y

Commutative property

y[n] = x[0] h[n] + x[1] h [n-1] + x[2] h[n-2] + x[3] h[n-3]

- k k]-n [ x [k]h [n]y

This will give the output y [n] as:

y[n] = x[n+1] h[-1] + h[1] x [n-1]

The highest sample point will be sum of

highest sample points of two signalsTotal no. of samples will be the

sum of samples in both the

signals -1. That is N1 + N2 -1

Examples on Discrete

Time Convolution

n

h[n] =u [ n ]

0

h [ n ] = u [n]

k

h [ 0 - k ]

0

k

u [ 1 - k ]

0

k

u [-1 – k ]

0

n0 1 2

x [ n ]

- k

]k -n [h ]k [ x [n]y

Here h[-k] is the mirror image of h [n]. And h[-

1-k] is the value for n= - 1 and h [+1-k] for n=1

Example Convolved Output for x[n] = n and h[n] = u[n]

We see that output is 0 for n < 0 and we get

output for n 0. So the summation is;

[n]u α - 1

α - 1 α [n]y

1nn

0 k

k

The output vale will start from zero and for

k the value saturates at ( 1 / ( 1 - )

x [ n ] = n Where 0 < < 1

For k and

If = 1 /2 than y[n] = 2

k

h [ - k ]

0

k 0-1-2

x [ k ]

Convolved Output for x[n] = 2n u[-n] and h[n] = u[n]

0

- k

k 2 [n]y

If we change the variable k = -m. We get the

integral from 0 to . This is when we are

computing only for n ranging from - to 0

2 1/2) - 1 (

1

2

1 [n]y

0 k

k

However we need to compute

the integral in the entire range

of “n” for which we get:

n

- k

k 2 [n]y If k we replace k by k = - k

n - k

k 2

1 [n]y If k we replace k by

k + n = m, we get

2

1 2

2

1 [n]y

0 m

m

n

0 m

n-m

1 n2 [n]y or

x[k] starts the output at - and

h[k] leads the output until . So

the output is from - to

Now considering the complete output from -

to n, where will also tend to infinity

The summation taken from - to 0 will show

that the output it must saturate at n=0

n

h [ n ]

0

h [ n ] = u [ - n]

k

h [ -k ]

0

k

u [ 1 - k ]

0

k0 1 2

x [ k ]

- k

]k -n [h ]k [ x [n]y

Here h[-k] is the mirror image of h [n]. And h[-

1-k] is the value for n= - 1 and h [+1-k] for n=1

Example Convolved Output for x[n] = n and h[n] = u[-n]

We see that output is 0 for n < 0 and we get

output for n 0. So the summation is;

[n]u α - 1

α - 1 α [n]y

1nn

0 k

k

The output vale will start from zero and for

k the value saturates at ( 1 / ( 1 - )

x [ n ] = n Where 0 < < 1

k0

Example of Convolved Output

x[n] = 1 for 0 n 4

= 0

h[n] = n 0 n 6

= 0 elsewhere

- k

]k -n [h ]k [ x [n]y

The convolved output will be from n=0 to n=10 and

the total no. of samples will be: (5+7 ) -1 = 11

Region #1: n < 0 y[n] = 0

n0 1 2

x [ n ]

3 4

n

h [ n ]

0 1 2 3 4 5 6

The starting sample will be n=0

The stop sample will be n=10

This is because of the fact that there is no overlap

So the convolved output will have 11 samples

Example Convolved Output where x[n] is of finite duration

x[n] = 1 for 0 n 4

= 0

- k

]k -n [h ]k [ x [n]y

In this case x[n-k] will keep moving below h[k] and

the complete fill up will be obtained for n = 4. In this

region 0 ≤ n ≤ 4. In this region k will also vary

from 0 to 4. In this region if n=1, than we summing

over n from 0 to 1 for the product of x [ n-k] and h

[k].

Region #2: n = 0 to n= 4. That

is the region 0 n 4.

n

0 k

k -n α [n]y

n

0 r

rα [n]y

α1

α1 [n]y

1n

n0 1 2

x [ n ]

3 4

n

h [ n ]

0 1 2 3 4 5 6

h[n] = n 0 n 6

= 0 elsewhere

If we put n – k = r. For k=0 : n = r and for k =n :

r = 0. So the summation on “r” changes from n

to 0. Which is same as r changing from 0 n. It is

like u[n] * h[n].

Here we will get y[n] for n=0 to n=4

k

h [ k ]

0 1 2 3 4 5 6

Visualization of Region #3

k

x [n -k ]

4 3 2 1 0

This is the where

region 2 ends

This is where

region 3 starts

This is the where region 3

stops. Last sample of h[k]

still overlaps

In this region 4 < n ≤ 6. That is the signal h [k]

remains completely filled only for two values of n,

which are n=5 and 6. But during this period the

summation for each value of n is to be carried for

all values of “k”, which varies from 0 to 4.

In region 3In this region 4 < n ≤ 6. The summation integral will be

for k= 0 to k = 4. and the output will be for n=5, and 6

4

0 k

k -n α [n]y

Here we get output for y[n] for n =5 and 6

4

0 k

k -n α α [n]y

) / 1 ( - 1

) /1 ( - 1 α [n]y

5n

k

x [n -k ]

4 3 2 1 0

k

x [n -k ]

4 3 2 1 0

k

h [ k ]

0 1 2 3 4 5 6

Visualization of Region #4

region 3

stops

In this region 6 < n ≤ 10. That is the signal x [n-

k] starts coming out. So far y[n] has reached at

y[6], making the output as 7 samples

In region 4: This is the region we are considering

n > 6. Now for n-6 ≤ k, the fist sample will come

out for n = 7 and the output will be obtained until

all sample get out that is upto k+4. So the value of

k will vary from : n-6 ≤ k ≤ 4

4

6-n k

k -n α [n]y

Region 5 we get when n > 10 than the output is zero

n - 10

0 r

6 r - α [n]y ) / 1 ( - 1

) /1 ( - 1 α [n]y

n - 11 6

k

x [n -k ]

4 3 2 1 0

Now if we put : k – (n – 6 ) = r The

limits of summation are from r=0

to r= 10 – n . And n – k = - r + 6

n - 10

0 r

r - 6 α α

Correlation / Correlation

Between Two Signals(continuous time)

f1()

f2()

T

∫-∞

∞

f1 ( ) f2 ( ) d

Definition of Correlation Generally Followed

The above integral show that there is no similarity between two signals. For finding similarity

we shift the waveform f2() towards left by time “t”. This shift is towards left which is negative

axis. So we take the value of “t” to be negative and integrate over “ “ . This is contradictory

to the convention of giving the leading shift to f2(t). In this case we perform the integral over

for a given shift of time “t”, which provides the value of integral

∫-∞

∞

f1 ( ) f2 ( - t) d 12 ( t ) =

12 (t) is defined as: f1() is stationary we advance f2()

by towards left by different “t” and compute the integral.

f2()

T

t

The sign on the shift in “t” is determined by the direction

of the axis (rather than delay or lead behaviour of signal)

in this case moving towards left is –ve axis. Here the

search parameter is “t”

Example of Correlation of two pulses

f1 ()

1

0 1

f2 ()

1

2 3

Correlation

This is in reference with the

starting point of the signal

f2()The term “t“ is –ve if we

move along the -ve direction of X-

axis. This automatically decreases

the delay.

t1 32

12 ( t )

This is same as:

∫-∞

∞

f1 ( ) f2 ( - t) d 12 ( t ) =

∫-∞

∞

f1 (x+t ) f2 (x) d x12 ( t ) =

∫-∞

∞

f1 (+t ) f2 () d 12 ( t ) =

The variable x can be replaced by This states that signal f2() can be kept

constant and we can move f1() towards

right side, which is positive direction of x-

axis. Here also for every shift the integration

is taking place on for different values of “t“

.

Correlation of a Returned Echo to a Radar

∫-∞

∞

f1 ( ) f2 ( - t ) d12 ( t ) =

That means we move f2( ) towards left for different values of time t ,

called as the scanning parameter. Since we are moving along the –ve

direction of x-axis the value of t is taken to be negative. In this process we

get the integral value for a given shift of time “t”

Cross-correlation Functions Similarity

∞

∫-∞

f1 ( t + ) f1 ( + T) d 11 ( t ) =

∫-∞

∞

f1 ( ) f2 ( - t) d12 ( t ) =

∫-∞

∞

f1 ( t + ) f2 ( ) d12 ( t ) =

∞

∫-∞

f1 ( t + k –T) f1 ( k) dk11 ( t ) =

Replace + T = k

= 11 ( t - T)

If f2(t) = f1 (t + T)

Signal Comparison : Correlation

∫-∞

∞

f1 ( ) f2 ( - t ) d12 ( t ) =

dthxtncorrelatioCross

dthhtncorrelatioAuto

xh

hh

)()()(

)()()(

Similarity Between Correlation and Convolution

Correlation Integral : ∫-∞

∞

x () h ( - t) dx h ( t ) =

∞

∫-∞

x () h { - (t - ) } d = x(t) * h (- t) x h ( t ) =

∞

∫-∞

x () h ( t - ) d x h ( t ) =

This can be written as:

∞

∫-∞

x () h ( - ( t - ) dx h ( t ) =

This can be written be interpreted as correlation integral of

x(t) and h(-t) which is nothing but convolution integral

The convolution integral can be written as:

t

f1(t)

t

f2(t)

T

∫-∞

∞

f1 ( ) f2 ( ) d

Correlation of following the Convention of Signal Theory

The other method of representing the correlation is

following the concept of signal processing concept.

That is shift to the left is lead of signal which has to be

positive and shift towards the right side has to be

negative identifying the lag in the signal. Under these

condition revised definition of correlation is:

∫-∞

∞

f1 (t ) f2 (t + ) d t12 ( ) =

. Both the conventions are valid. One

can follow either of the convention

Continuous Time

Convolution

Defining a Pulse for Zero Order Hold

If we multiply by T(t) to any signal x(t) and it is a sample and

hold circuit. Than it will hold the value of x(t) at x(0) for a

duration t +T, and it will also reduce the sample value by 1 /T.

Therefore to preserve the original sample value at t = 0, we

need to multiply the x(t) by the value T. So

t

T(t)

1/T

T

T(t) = 1/T for 0 t T

Otherwise0

We are defining the pulse as:

x /(0) = x ( 0 ) T ( t ) T

x /(1) = x ( T ) T ( t –T ) T

x /(2) = x ( 2T ) T ( t -2T ) T

T kT)-(t (kT)x (t)x-k

T

/

We can approach to x ( t ) from x/ (t) in

the limiting case T tends to zero

x /(-1) = x ( -T ) T ( t + T ) T

t

T(t) 1/T

T


T

/

When T 0 than we are approaching very

close to the original signal. The physical

interpretation is that we are passing the

signal through a wide band bandwidth. So we

will get the original signal.

Representation of continuous time signal

in terms of Impulse Zero Order Hold

(t) x T kT)-(t (kT)x (t)x-k

T0 T lim

/

t

x ( t )

T 2T0 3T-T

Further as T 0 than the function x /(t)

x(t) and summation approaches to

integral

Defining a Pulse for Zero Order Hold

Originally the pulse was T (-), which is shifted towards

right side by “t” such that the centre of pulse is integer

multiple of T. That is = mT, where m is an integer

T( - )

1/T

- T

T(t - )

1/T

- T tt –T

mT

Shifted towards right side by time “t”

T() 1/T

T

T(t) = 1/T for 0 t 1/T

Otherwise0

We are defining the pulse as:

lim𝑻 →𝟎

𝜹𝑻 𝒕 → 𝜹 (𝒕)

Sampling the signal x ( ) at T (t - )

T(t - )1/T

tt –T

mT

x ( )T(t - )

Shift in T ( ) by “t”

Which is sample value of x ( ) at t =

mT. Because the pulse duration is T

and the amplitude is ( 1 /T).

d ) - t ( ) (x (t)x -

Under this condition we value of

x(t) corresponding to m = k. Other

values are zero. That means we get

the value of x(t) for t =

(t) x


T0 T lim

/

The shaded region has the area:

x (mT) x T x ( 1 / T) = x (mT)

The area under the curve x ( ) T(t - )

will also approach to x( m T) is T 0.

In the limiting case as T 0 the function


T(t - )1/T

- T tt –T

mT

x ( )T(t - )

Shift in T ( ) by “t”The shaded region has the area:

x (mT) x T x ( 1 / T) = x (mT)

x (mT). Which is sample value of x ( ) at

t= mT. Because the pulse duration is T and

the amplitude is ( 1 /T).

The area under the curve x ( ) T(t - )

will also approach to x( m T) is T 0.

d ) - t ( ) (x (t)x -

While finding the area if we consider

linear variation from t – T ≤ ≤ t, We get

the sampled value of x( ) ( t--T/2). This

the value at the centre. This value

approaches to x(mT). For this value of x()

is x (mT) when it is sampled by a pulse

T(t-mT). Further the pulse T(t) also

becomes an impulse if T 0

(t) x


T0 T lim

/

Continuous Time Unit Step and Unit Impulse Function

t

u ( t )

1

h ( t ) = u(t)

Ideal Unit

Step Function

t

0

aτ - τd e (t)y

x ( )

e – a t

u(t)

x( t ) = e – a t

u(t)

u ( - )

1

Region #1

The output will increase as a

function of t from 0 < ≤ t

As “t” tends to infinity y(t) = ( 1 /a)

Region #1

t

0 τ

τa - | ) e ( ) a

1 - ( (t)y

) e - 1 ( ) a

1 ( (t)y ta -

For all values of t the y(t) = ( 1 /a) ( 1 – e –a t) u(t)

Continuous Time Unit Step and Unit Impulse Function

t

u ( t )

1

h ( t ) = u(t)

Ideal Unit

Step Function

t

aτ τd e (t)y

x( t ) = e a t

u( -t)

u ( - )

1 Region #1

The output will increase as a

function of t from - < ≤ t

Region #1

t

τ

τa | ) e ( ) a

1 ( (t)y

t

x( t ) 1

e a t

u ( -t)

y (t ) = ( 1 /a) e at for t ≤ 0

Region #2

The output will remain constant

as function of t from < 0 ≤ t

Region #2

y ( t ) = 1 /a


d ) - t ( ) (x (t)x -

This represents the convolution of

x(t) with impulse (t)

In case x(t) is u(t) than we get the integral as: d ) - t ( ) (u (t)u -

d ) - t ( (t)u 0

Properties of Systems

Memory and

Memoryless Systems

System Without Memory

A system is said to be memoryless, if its output at any instant of time

Dependents only the value of input at that instant of time. That is y[n]only

depends on the input x[n].

This is only possible when h [n] exists only at n=0 otherwise it is zero.

Such an impulse response could only be unit step function [n]. The

amplitude of [n] may not be unity. It may have the arbitrary value k.

x [ n ]K [ n ]

y [ n ] = k x [ n ]

x [ n ]K h [ n ]

y [ n ] = k ( x [ n ] * h[n] )


y [ n ]x [ n ]K [ n ] y [ n ] = k x [ n ]

]k -n [ [k]k x [n]y -k

If k=1 than it becomes an identity system

d ) - t (h ) (x x(t)-


A system is said to be memoryless, if its output at any instant of time

Dependents only the value of input at that instant of time. That is y[n]

only depends on the input x[n].

Such a system is called as identity system. Because the input is same as output

This is being visualized as if the input is [n] and system is characterized by

The impulse response x [n]. For such a system the output will be x [ n ]

y [ n ]x [ n ]h [ n ] y [ n ] = x [ n ] * h [n]

This is only possible when the input x [n] = [n]

y [ n ] = h [ n ] * x [n]y [ n ]h [ n ]

x [ n ]

It is analogus to

Identity Systems

That means h [ n ] = [n]

For an identity system h [ n ] = [ n ]

n

-k

]k -n [ [k]x [n]x

And

-

τd ) τ- t ( δ ) τ( x (t)x

y [ n ] = [ n ] * x [n]

y [ n ] [ n ]x [ n ]

System With Memory

1-n

-k

[n] x [n]x [n]y

The summation term gives the

accumulated sum to which the

Current value x [n] is added

Discrete Time: An accumulator

In continuous time: Capacitor Charging

An example of a system with the memory is charging of capacitor. The

capacitor charges to the previous value.

t

-

τd ) τ( x C

1 (t)y

The charging of a capacitor by an

Input current x(t) keeps charging

The capacitor to a voltage y(t)

An example of memory system is an accumulator. It adds

the current value to the previous values

n

-k

[n]x [n]y

[n] x ] 1 -n [y [n]y

System With Memory

y [n] = x [n -1]Here the value of y [n] at any instant n = no

depends on the earlier value of x [n]. System

With memory

A system with memory stores the information about the value of input value

other than current time. For example a delay must retain the previous .

Here x[n] already exists and we give a unit delay than it becomes y [n]

y [n] = x [n -1]x [n]

Here y [n] is response to the signal which

was existing. Therefore the system is causal

y [n] = x [n-1]

DELAY

x [n]Delay

Invertibility / Inverse of Systems

Invertible System / Inverse System

A system is invertible: Than an inverse system exists, which when cascaded

with the original system yields the output, which

is same as the input

y [n] = 0 It is a noninvertible system because the output is zero for all inputs

y (t) = x2 (t) is a noninvertible system, because we can not determine

determine the sign from the knowledge of output signal

Invertibility and Inverse System for Continuous Time

A system is said to be INVERTIBLE if distinct inputs lead to distinct output

If a system is invertible than a inverse system exists that, when cascade

With the original system yields the output, which is basically input signal

Continuous Time System

y(t) = 2 x(t)x ( t )

y ( t ) w ( t )

w ( t ) = x ( t )21w (t) = y (t)

systemx ( t ) y ( t ) w ( t )

w ( t ) = x ( t )Inverse system

Invertibility and Inverse System: Discrete time



Discrete Time System

x [ n ] y [ n ]

w [n] = y [n] – y [ n – 1 ]

w [n]

[k] x [n]y n

- k

[n] x [k] x [k] x ]n [y 1 -n

0 k

n

- k

[n] x ] 1 -n [y [n]y

w [n] = y [n] – y [ n – 1 ]

Difference Equation

Integrator

Three basic Operators to Perform a Function in Discrete Time

[n] x b ] 1 -n [y a [n]y

y [n] a y [n] a

multiplier

y [n] y [n -1] DelayDelay

+y [n]

y [n-1]

y[n] + y [n-1] Summation

The delay element can be considered as a simple RS flip-flop

Discrete time Accumulator




x [ n ] y [ n ] [k] x [n]y

n

- k

[n] x [k] x [k] x ]n [y 1 -n

0 k

n

- k

[n] x ] 1 -n [y [n]y

First system can be obtained by passing x[n]

through u[n] and its system representation is shown

System Implementation of Discrete Time Accumulator

Convolution of x[n] with h[n]


1-n

-k

[n] x [n]x [n]y

n

-k

[n]x [n]y [n] x ] 1 -n [y [n]y

So this can be expressed as :

-k

]k -n [u [n]x [n]y

n

-k

]k -n [u [n]x [n]y

h[n] = u [n]x [n] y [n]

y [n] = x[n] * u[n]

In accumulator we add present value in the previous value. y[n] = y [n-1] + x[n]

System Implementation of Discrete Time Accumulator

Discrete Time Accumulator System

n

-k

[n]x [n]y

[n] x ] 1 -n [y [n]y

-k

]k -n [u [n]x [n]y

h[n] = u [n]x [n] y [n]

y [n] = x[n] * u[n]

Unit

Delay

x [n] y [n]

y [n - 1]

Integrator should have feed-back. That is what is reflected

Impulse Response of Discrete Time Accumulator

Unit

Delay

x [n] y [n]

y [n - 1]

If we assume system to be

causal. That is y[n-1]=0. And the

input is [n]. Than the output

will be u [n]

y [n] = y [ n – 1 ] + x [n]

Input is [n]: Output is u[n]:

If we multiply the feed-back by “a” after the delay. The

output will be: 1 + a + a2+ ….. The output is: y [ n ] = an u[n]

The system has recursion. Therefore, we called as the

recursive system. Further the impulse response is infinite.

So we call it as infinite Impulse Response System

Recursive Response : I I R : all Pole System

The output depends on present

and past values of inputx [ n ] b0

D-a1

-a2

D

-ak

D

y [ n ]

The difference equation is:

y [n] = b0 x [ n ] - a1 y [ n – 1 ] - a2 y [ n – 2 ] - ….. - ak y [ n – k]

N

1k

0k ]n[xb]kn[ya]n[y

y [n] = y [ n – 1 ] + x [n]

If we consider that coefficients ak and b0 are normalized value w.r.t.

to a0. Than the above equation can be written as:

N

0k

0k ]n[xb]kn[ya

Difference Equation

Differentiator

Invertibility and Inverse System: Discrete time

The system is invertible if the output of accumulator

is fed through a first order differentiator. Here the

second block is the differentiator


x [ n ] y [ n ]

w [n] = y [n] – y [ n – 1 ]

w [n]

[k] x [n]y n

- k

[n] x [k] x [k] x ]n [y 1 -n

k

n

k

1] -[n y ]n [y ]n [x

Impulse response of a differentiator

and invertible System

The impulse response of the differentiator is: [n] - [ n – 1]

Discrete Time Differentiator

x [ n ]

y[n] = x [n] – x [ n – 1 ]

y [n]

Impulse response of an integrator and differentiator

u [n] * { [n] - [ n – 1]} = u [n] [n] – u [n-1 ] [ n – 1]}

u [n] - u [n-1 ] = [ n ]

This establishes the invertiability of the system

System Implementation of

Differentiator

Discrete Time Differentiator

x [ n ]

y[n] = x [n] – x [ n – 1 ]

y [n]

y[n] = x [n] – x [ n – 1 ]

y [n] = x [n] – x [ n – 1 ]

Unit

Delay

x [n]

The impulse response such as system is : [ n ] - [ n -1]. Since it is a feed-forward

So the response will be of finite duration. These are non-recursive systems

Differentiator: Non recursive System

y[n]

y [n] = x [n] – x [ n – 1 ]

Dx [n]

The impulse response such as system is : [ n ] - [ n -1].

Since it is a feed-forward .So the response will be of finite

duration. These are non-recursive systems

D

D

bo

-b1

-bk

-b2

y [n] = box [n] + b1 x [ n – 1 ]+ b2 x [ n – 1 ]+…… bk x [ n – k ]

]kn[xb [n]y k

M

0k

System With Memory: Discrete Time




n

-k

[n]x [n]y [n] x ] 1 -n [y [n]y

This is feed-forward so it is no-recursive type of system

Difference Equation

y[n] = x [ n ] – x [n-1]

This equation has the feed-back. This reflects the recursive nature

Impulse Response of a Reversible System

Discrete Time Accumulator

n

-k

[n]x [n]y

[n] x ] 1 -n [y [n]y

Verification of Associative Property

Differentiatory[n] = x [ n ] – x [n-1]

Unit

Delay

x [n] y [n]

y [n - 1] h[n] = u [n]

Unit

Delay

x [n] h[n] = [n] - [n-1]

y[n] = u [ n ] * { [n] - [n-1] }

y[n] = { [n] - [n-1] } * u [n]

General Form of a Discrete Time System

It is a combination of a accumulator and differentiator

n

-k

[n]x [n]y y[n] = y[n-1] + x[n]Accumulator

y[n] = x [ n ] – x [n-1]DifferentiatorAs such both are called as first

order difference equation

D

x [n] b0

- b1

y [n]

D

- a1

y1[n] = b0 x[n] – b1 x[n-1]

y1 [n]

Differentiator

Integrator

y [n] = y1 [n-1] – a1 y[n-1]


y[n] + a1 y[n-1] = y1 [n] = b0 x[n] – b1 x[n-1]

If we consider all the coefficients were initially normalized to some

constant coefficient a0. That means a1 was actually a1/a0, b0 was b0/a0

andb1 was b1 / a0

M

0k

k

N

0k

k ]k -n [y b ]k -n [y a

D

x [n] b0

- b1

y [n]

D

- a1

y1[n] = b0 x[n] – b1 x[n-1]

y1 [n]

Differentiator

Integrator

y [n] = y1 [n] – a1 y[n-1]

Inverse System for Continuous and Discrete Time

A system is said to be INVERTIBLE if the complete system functions

As an Identity system

h1 (t)x ( t ) y ( t ) w ( t )

w ( t ) = x ( t )h2 ( t)

h1 (t) * h2 ( t ) = (t)

h1 [n]x [ n ] y [n] w [ n ]

w [ n ] = x [ n ]h2 [ n ]

h1 [ n ] * h2 [ n ] = [ n ]

Inverse System for Continuous Time

A system is said to be INVERTIBLE if the

complete system functions As an Identity system

h1 (t)x ( t ) y ( t ) w ( t )

w ( t ) = x ( t )h2 ( t)

y ( t ) = x (t) * h1(t) = x ( t – to )

That means y (t) is nothing but delay x (t) by time t0.

h1 (t)x ( t )

y ( t ) = x ( t – t0) y ( t ) = x ( t – t0)

If y (t) = x (t) than it is a memoryless system and the impulse response

Has to be (t). Now x ( t - t0 ) is x(t) shifted by t0. That means the impulse

response is ( t - t0 ), this is because that the output now exists at t = to

Inverse System for Continuous Time

A system is said to be INVERTIBLE if the complete system functions

As an Identity system

h1 (t)x ( t ) y ( t ) w ( t )

w ( t ) = x ( t )h2 ( t)

y ( t ) = x ( t – to ) h ( t ) = ( t – to)

If t – to > 0 than the output at time “t” is what the input is at time t - to

To get w( t ) = x( t ), we should advance the response y (t) by time to

That means the impulse response h1(t) = ( t + to )

h1( t ) * h2 ( t ) = ( t – to ) * ( t + to ) = ( t )

Inverse System in Discrete Time: Identity System

h1 [n]x [ n ] y [n] w [ n ]

w [ n ] = x [ n ]h2 [ n ]

y [ n ] = x [ n – no ] h1 [ n ] = [ n – no ]

w [ n ] = y [ n+ no ] h2 [ n ] = [ n + no ]

k

[ n - k + no ]

0 no

-k

oo ]n k -[n ]-n[k [n]h

k

[ k – no ]

no0

h1 [k]

k

[ k + no ]

no0

h2 [k]

Inverse System in Discrete Time: Identity System

h1 [n]x [ n ] y [n] w [ n ]

w [ n ] = x [ n ]h2 [ n ]

h [ n ] = h1 [ n ] * h2 [ n ]

= [ n - no] * [ n + no] = [n]

y [ n ] = x [ n – no ] h1 [ n ] = [ n – no ]

w [ n ] = y [ n+ no ] h2 [ n ] = [ n + no ]

k

[ k – no ]

no0

k

[ n - k + no ]

0 no

-k

oo ]n k -[n ]-n[k [n]h

Stability of the System

Stability for LTI System : Discrete Time

A system is said to be stable if bounded input has the bounded output

Let the input x [ n ] is a bounded in magnitude. | x ( t ) | < B for all values of n

| τd ) τ- t ( x ) τ(h | | ) t (y | -

τd | ) τ- t ( x | | ) τ(h | | ) t (y | -

Since, | x ( t ) | < B for all values of t

τd | ) τ(h | B | ) t (y | -

That means for a system to be stable

τd | ) τ(h | -

Stability for LTI System: Continuous Time

A system is said to be stable if bounded input has the bounded output

Let the input x [ n ] is a bounded in magnitude. | x [ n ] | < B for all values of n

| ]k -[n x | | [k]h | | ]k -[n x [k]h | | [n]y |-k-k

-k

| [k]h | B | [n]y |

Since | x [ n – k ] | < B for all values of k and n

This requires that the impulse response is absolutely summable

| [k]h | -k

for all values of n

We take the magnitude because the swing +ve and negative values may tend to infinity

but still the average value may be zero. This summation may be zero

Time Invariance Systems

Time Invariance System

A system is said to be time invariant if the behavior and

characteristic of the system are fixed over time.

A system is said to be time invariant if the time shift in the

Input signal results in time shift in the output signal

In a continuous time system if y (t) is the output for an input x (t)

Than y ( t – t0 ) is the output when input is x ( t – t 0 )

For a discrete time system if y [n] is the output for an input x [n]

Than y [ n – n0 ) is the output when input is x [ n – n 0 )

If we take a R, C circuit and the values of the elements do not change with

time (characteristic remains fixed with time). It is a time invariant system. If

the values of the elements changes with time than it is time variant system.

Checking a System for Time Invariance

Let us consider a system y ( t ) = sin [ x1 ( t ) ]

Let x1 (t) is the input at any time t y1(t) = sin [ x1(t) ]

Now if we look at the response delayed time t0.

That is for the input x1( t - t0 )

x2 ( t ) = x1 ( t – t0) Y2 ( t ) = sin [ x1 ( t – t 0 )

Now if we find y1 (t) at time = t - t0y1( t – t0 ) = sin [ x1( t – t0) ]

So the system is time invariant

The time invariance should be checked for any input and any time delay.

Some Useful Signal Operation : Time Scaling

t

f1 (t)

T1

2

T2

2

Case (1)

f1 ( t) = g (2 t)

For a > 1 there is

A time compression.

In the present case a = 2

t

f2 (t)

2T1 2T2

Case (2)

For a < 1 there is

A time expansion.

In the present case

a = 0.5

Original Signal g ( t )

t

g (t)

T1T2

Signal f ( t ) = g ( a t )

g (a t )

t = T /a

= g ( T )

Checking a System for Time Invariance

t

x1 (t)1

- 2 2

y ( t ) = x ( 2 t )

t

y1 (t)

-1 1

Time compression

x2 ( t ) = x1 ( t – 2 )

t

x2 (t)1

4t

y2 (t)

2

y1 ( t ) = x1 ( 2 t )

y2 ( t ) = x1 ( 2 t - 2 )

Since we have delayed x1 ( t ) by 2, we

would expect that y1 ( t ) should also be

delayed by 2 and output should be:

t

y2 (t)

1 3

Expected from

Time invariant

system

So the system is not time Invariant

Any time shift also leads time compression in output. So it is a time VARIANT SYSTEM

Causality of Systems

Causality of a System

A non-anticipative system is called as Causal System

Output depends on present and past values of input signals.

h(t) = 0 for t <0

A typical is charging of a capacitor. The voltage on capacitor

Depends on present past and past values of source voltage

h [n] = 0 for n <0

For a discrete time system

For a continuous system

Causality of a System for discrete System



Therefore the criteria of present and past value of x[n] states that

for any shift in x[n] the h[n] must shift so:

]k -[n h [k]x [n]y -k

n

Since h [n] , for n < 0. The above expression can also be

written in alternate form by using commutative property

]k -[n h [k]x [n]y 0k

h [ n ] = 0 for n < 0

Causality of a System for Continuous Time System



Therefore the criteria of present and past value of x ( t ) states that

for any shift in x (t) the h ( t ) must shift so:

Since h ( t ) , for t < 0. The above expression can also be

written in alternate form by using commutative property

h (t) = 0 for t < 0

τd ) τ- t (h ) τ( x ) t (y

t

-

τd ) τ(h ) τ- t ( x ) t (y

t

-




The system y [ n ] = x [ n – 1 ] is causal system

The system y ( t ) = x ( t – t0 ) is causal System

A typical is charging of a capacitor. The voltage on capacitor

Depends on present past and past values of source voltage

If a signal is defined at t = to. Than we can determine the output

at any instant later t > to.




The system y [ n ] = x [ n ] + x [ n + 1 ] is NON causal system

The system y ( t ) = x ( t + t0 ) is NON Causal System

Any memory less system is Causal system, because it depends

Only current value of the input signal

Checking Causality of a System

y [ n ] = x [ - n ]

The values at n = n0 for some positive time n0 depends on past

values of signal x [ - n0 ]. System seems to be Causal

But if n is negative say n = - 4. Than y [ - 4 ] = x [ 4 ]. That means

output depends on a future value of input signal. So system is not Causal

y ( t ) = x ( t ) cos ( t + 1 )

Here cos ( t + 1) indicates that system is not causal

But the output is multiplications of x(t) which defines

That the system is casual

Causality not an Essential Parameter

If Time is not Independent Parameter

This finds application in meteorological signals, predicting

the stock Market. In such cases it is the average value taken

over an interval say – M to M and predict the value at some

instant of time

M

Mk

M][kx 12M

1[n]y

This can be used to predict the dynamic behavior f the market

Linear Constant

Coefficient Difference

Equation

Discrete Time


[n] x b ] 1 -n [y a [n]y

y [n] a y [n] a

multiplier

y [n] y [n -1] DelayDelay

+y [n]

y [n-1]

y[n] + y [n-1] Summation

The delay element can be considered as a simple RS flip-flop

In this the information Q = 1 gets tranf





n

-k

[n]x [n]y [n] x ] 1 -n [y [n]y

This is feed-forward so it is no-recursive type of system

Difference Equation

y[n] = x [ n ] – x [n-1]

This equation has the feed-back. This reflects the recursive nature


It is a combination of a accumulator and differentiator

n

-k

[n]x [n]y y[n] = y[n-1] + x[n]Accumulator

y[n] = x [ n ] – x [n-1]DifferentiatorAs such both are called as first

order difference equation

y[n] + a1 y[n-1] = b0 x[n] +b1 x[n-1]

D

x [n] b0

- b1

y1 [n]

D

- a1

y [n]

y1[n] = b0 x[n] + b1 x[n-1]

y[n] = y1[n] - a1 y[n-1]


D

x [n] b0

- b1

y [n]

D

- a1

If we consider all the coefficients were initially normalized to some

constant coefficient a0. That means a1 was actually a1/a0, b0 was b0/a0

andb1 was b1 / a0

M

0k

k

N

0k

k ]k -n [y b ]k -n [y a

y[n] + a1 y[n-1] = b0 x[n] +b1 x[n-1]

N th Order linear constant Coefficient Difference Equation

M

0 k k

N

0 k k ]k -n [ x b ]k -n [y a

M

0 k k

N

1 k k0 ]k -n [ x b ]k -n [y a [n]y a

This equation can be expressed as:

} ]k -n [y a - ]k -n [ x b { ) a

1 ( [n]y

N

1 k k

M

0 k k

0

This equation states that for computing the present output value. We need

the information of the previous output value. From this we must realize

that we need the auxiliary conditions. That is the initial conditions.

To compute y [n] we need the information on y [n-1], ……..y [ n – N]

N th Order linear constant Coefficient Difference Equation

M

0 k k

N

0 k k ]k -n [ x b ]k -n [y a

In this equation if we assume that N=0. That means

K =0. For this the above equation gets modified as

M

0 k 0k ]k -n [ x )a / b ( [n]y

M

0 k k0 ]k -n [ x b [n]y a or

In this case the output depends on the present and past (previous)

values of the input. That means we do not recursively use previously

computed values of the output to compute the present value of output

N th Order linear Constant Coefficient Difference Equation

M

0 k 0k ]k -n [ x )a / b ( [n]y

M

0 k

]k -n [ x k) h( [n]y

Now if we compare the above two

equations we find that the impulse

response of the system is:

h [n] = bn / ao for 0 n M

0 otherwise

The above equation is convolution sum of the input x [n] with the impulse

response ( bn /ao). Since we are not using the previously computed value to

compute the present value, therefore we do not need the auxiliary

conditions. That is the status of the system whether It is initially at rest.

PRACTICAL EXAMPLE

y [n] = ( 1 / 2 ) y [n-1 ] + x [n]

Now we want to find the output for x [n] = K [n]. This

forces the condition that x [n ] = 0 for n -1. The

condition of initial rest implies that y [n] = 0 for n -1

In the process we take the Causal LTI system. The causal system

dictates that x [n] = 0 for n < 0. Identically y[n] = 0 for n<0

From the above equation we can y[n] for an input x [n] = K [n]

y [0] = ( 1 /2 ) y [-1] + k [n] = k

y [1] = ( 1 /2 ) y [0] + x [1] = ( 1 / 2 } k

PRACTICAL EXAMPLE

y [n] = ( 1 / 2 ) y [n-1 ] + x [n]

Now we want to find the output for x [n] = K [n].

This forces the condition that x [n ] = 0 for n -1.

The condition of initial rest implies that y [n] = 0 for n -1

In the process we take the Causal LTI system. The causal system

dictates that x [n] = 0 for n < 0. Identically y[n] = 0 for n<0

From the above equation we can y[n] for an input x [n] = K [n]

y [0] = ( 1 /2 ) y [-1] + k [n] = k

y [1] = ( 1 /2 ) y [0] + x [1] = ( 1 / 2 } k

PRACTICAL EXAMPLE

y [n] = ( 1 / 2 ) y [n-1 ] + x [n]

y [0] = ( 1 /2 ) y [-1] + k [n] = k

y [1] = ( 1 /2 ) y [0] + x [1] = ( 1 / 2 ) k

y [2] = ( 1 /2 ) y [1] + x [2] = ( 1 / 2 ) 2 k

y [3] = ( 1 /2 ) y [2] + x [3] = ( 1 / 2 ) 3 k

So y [n] = ( 1 / 2) n k

Since the system is completely characterized by an impulse response

The impulse response can be found from putting k=1

So h [n] = ( 1 / 2 ) n u[n]

Linear Constant

Coefficient Difference

Equation

Discrete Time

Miscellaneous Slides

Some Mathematical Functions

cos ( A + B ) = cos A cos B – sin A sin B

cos ( A - B ) = cos A cos B + sin A sin B

sin ( A + B ) = sin A cos B + cos A sin B

sin ( A - B ) = sin A cos B - cos A sin B

sin A sin B = 1 / 2 { cos ( A- B) - cos (A + B) }

cos A cos B = 1 / 2 { cos ( A- B) + cos (A + B) }

sin A cos B = 1 / 2 { sin ( A+ B)+ sin (A - B) }

cos A sin B = 1 / 2 { sin ( A+ B) - sin (A - B) }

sin2 = 2sin cos

cos2 = cos2 – sin2

Linear Constant Coefficient Difference Equation

Formation of Difference Equation an Impulse Response

Delay

x [ n ] y [ n ]

a y [ n -1 ]

a

y [n] = x [ n ] + a y [ n – 1 ]Now let us assume that it is a causal

system y [ n ] = for n < 0

The system response for [ n ] is:

y [0] = x [ 0 ] + a y [– 1 ] = 1 Since x [ n ] is [ n ] so x [ o ] = 1 and y [ 0 ] =1

y [1] = x [ 1 ] + a y [0] = a

y [2] = x [ 2 ] + a y [1] = a2Or y [n] = x [ n ] + a y [n-1] = an u [n]

The impulse response of the system is: a n

u [ n ]


and past values of input

Summation

x [ n ]

y [ n ]

b

k

D D D D

b0b1

b2

X ( z) = ∑- ∞

∞x [n] z

- n

X/( z) = ∑

- ∞

∞x [ n - 1 ] z

- n

X/( z) = z

– 1X [ z ]

z b ) z ( Y k -

k

M

o k

) z ( X ) z b .......... z b z b b ( -k

k

-2

2

1 -

1 o

) z ( X

) z ( Y

k

k

2-k

2

1 -k

1

k

o

Z

) b .......... z b z b zb (

Basically it is a all zero filter. It has all the poles lying at z =0. Origin

Non Recursive Response : F IR : all Zero System

Recursive Response : I I R : all Pole System


and past values of input

y [ n ] Y ( z)

) z ( X ) z a .......... z a z a (1 ) z ( Y -k

k

-2

2

1 -

1

Basically it is a all pole filter. It has all the zeros lying at z =0. Origin

Y [ n - 1 ] z – 1

Y ( z)

x [ n ] 1

D-a1

-a2

D

-akD

y [ n ]

The difference equation is:

y [n] + a1 y [ n – 1 ] + a2 y [ n – 2 ] + ….. ak y [ n – k] = x [ n ]

) z a .......... z a z a (1

1

) z ( X

) z ( Y

k-

k

2-

2

1 -

1

Time Integration Property

Problem 2: Addition of any constant to x (t) does not change the value of f (t).

That means for given f (t), x (t) can have many possible solutions.

That means we need to find the unique function x (t), which relates

uniquely the integral of function f (t).

LTI System with

The response f (t)

Input (t) ∫- ∞

t

y ( t ) = f ( ) ( t - ) d = f ( t )

LTI System with

the response f (t)

Input (t) f ( t )

∫- ∞

t

x (t) = f ( ) d Integrator

LTI System with

the response f (t)

Input (t) ∫- ∞

t

x (t) = f ( ) u ( t - ) d Integrator

u(t)

(Using associative property)

convolution discrete and continuous time-difference equaion and system properties (1)

Engineering