1 introduction chapter 1.6 elementary signals ★ 1.6.1 exponential signals (1.31) b and a are real...

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IntroductionIntroduction CHAPTER

1.6 Elementary Signals1.6 Elementary Signals

★ ★ 1.6.1 Exponential Signals1.6.1 Exponential Signals ( ) atx t Be (1.31)

B and a are real parameters

1. Decaying exponential, for which a < 02. Growing exponential, for which a > 0

Figure 1.28 (p. 34)(a) Decaying exponential form of continuous-time signal. (b) Growing exponential form of continuous-time signal.

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Ex. Lossy capacitor: Fig. 1-29Fig. 1-29.

Figure 1.29 (p. 35)Lossy capacitor, with the loss represented by shunt resistance R.

( ) ( ) 0d

RC v t v tdt

KVL Eq.:

(1.32)

/( )0( ) t RCv t V e (1.33)

RC = Time constant

Discrete-time case:

[ ] nx n Br (1.34)

r e

where

Fig. 1.30Fig. 1.30

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Figure 1.30 (p. 35)(a) Decaying exponential form of discrete-time signal. (b) Growing

exponential form of discrete-time signal.

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★ ★ 1.6.2 Sinusoidal Signals1.6.2 Sinusoidal Signals

( ) cos( )x t A t

2T

(1.35)

where

( ) cos( ( ) )

cos( )

cos( 2 )

cos( )

( )

x t T A t T

A t T

A t

A t

x t

periodicity

◆ Continuous-time case:

Fig. 1-31Fig. 1-31


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Figure 1.31 (p. 36)(a) Sinusoidal signal A cos( t + Φ) with phase Φ = +/6 radians. (b) Sinusoidal signal A sin ( t + Φ) with phase Φ = +/6 radians.

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Ex. Generation of a sinusoidal signal Fig. 1-32Fig. 1-32.

Figure 1.32 (p. 37)Parallel LC circuit, assuming that the inductor L and capacitor C are both ideal.

Circuit Eq.:2

2( ) ( ) 0

dLC v t v tdt

0 0( ) cos( ), 0v t V t t

(1.36)

(1.37)

where0

1

LC (1.38)

Natural angular frequency of oscillation of the circuit

◆ Discrete-time case :

[ ] cos( )x n A n

[ ] cos( )x n N A n N

2N m 2

radians/cycle, integer ,m

m NN

(1.40)

(1.39)

Periodic condition:

or

Ex. A discrete-time sinusoidal signal: A = 1, = 0, and N = 12. Fig. 1-33.Fig. 1-33.

(1.41)

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Figure 1.33 (p. 38)Discrete-time sinusoidal signal.

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Example 1.7 Discrete-Time Sinusoidal SignalA pair of sinusoidal signals with a common angular frequency is defined by

1[ ] sin[5 ]x n n 2[ ] 3cos[5 ]x n n

1 2[ ] [ ] [ ]y n x n x n

and

(a) Both x1[n] and x2[n] are periodic. Find their common fundamental period.

(b) Express the composite sinusoidal signal

In the form y[n] = Acos(n + ), and evaluate the amplitude A and phase .

<Sol.><Sol.>(a) Angular frequency of both x1[n] and x2[n]:

5 radians/cycle 2 2 2

5 5

m m mN

This can be only for m = 5, 10, 15, …, which results in N = 2, 4, 6, …(b) Trigonometric identity:

cos( ) cos( )cos( ) sin( )sin( )A n A n A n x1[n] + x2[n] with the above equation to obtain thatLet = 5, then compare

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sin( ) 1 and cos( ) 3A A

1

2

sin( ) amplitude of [ ] 1tan( )

cos( ) amplitude of [ ] 3

x n

x n

sin( ) 1A

1

2sin / 6

A

[ ] 2cos 56

y n n

= / 6

Accordingly, we may express y[n] as

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★ ★ 1.6.3 Relation Between Sinusoidal and Complex Exponential Signals1.6.3 Relation Between Sinusoidal and Complex Exponential Signals

1. Euler’s identity: cos sinje j (1.41)

Complex exponential signal: jB Ae (1.42)

cos( ) Re{ }j tA t Be (1.43)

( )

cos( ) sin( )

j t

j j t

j t

Be

Ae e

Ae

A t jA t

( ) cos( )x t A t (1.35)

◇ ◇ Continuous-time signal in terms of sine function:Continuous-time signal in terms of sine function:

( ) sin( )x t A t (1.44)

sin( ) Im{ }j tA t Be (1.45)

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/ 4/ 4


cos( ) Re{ }j nA n Be (1.46) (1.47) and

3. Two-dimensional representation of the complex exponential e j n for = /4 and n = 0, 1, 2, …, 7. : Fig. 1.34Fig. 1.34.

Projection on real axis: cos(n);Projection on imaginary axis: sin(n)

Figure 1.34 (p. 41)Complex plane, showing eight points uniformly distributed on the unit circle.

sin( ) Im{ }j nA n Be

2. Discrete-time case:

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★ ★ 1.6.4 Exponential Damped Sinusoidal Signals1.6.4 Exponential Damped Sinusoidal Signals

( ) sin( ), 0tx t Ae t (1.48)

Example for A = 60, = 6, and = 0: Fig.1.35Fig.1.35.

Figure 1.35 (p. 41)Exponentially damped sinusoidal signal Ae at sin(t), with A = 60 and = 6.

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1( )

tv d

L


Ex. Generation of an exponential damped sinusoidal signal Fig. 1-36Fig. 1-36.

Figure 1.36 (p. 42)Parallel LRC, circuit, with inductor L, capacitor C, and resistor R all assumed to be ideal.

Circuit Eq.:1 1

( ) ( ) ( ) 0td

C v t v t v ddt R L

(1.49)

/(2 )0 0( ) cos( ) 0t CRv t V e t t

0 2 2

1 1

4LC C R

(1.50)

where (1.51) /(4 )R L C

Comparing Eq. (1.50) and (1.48), we have

0 0, 1/(2 ), , and / 2A V CR

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◆ Discrete-time case:

[ ] sin[ ]nx n Br n (1.52)

★ ★ 1.6.5 Step Function1.6.5 Step Function


1, 00, 0[ ] n

nu n (1.53)

Figure 1.37 (p. 43)Discrete-time version of step function of unit amplitude.

Fig. 1-37.Fig. 1-37.

x[n]

n1 2 3 40123

1

1, 0( )0, 0

tu tt


(1.54) Figure 1.38 (p. 44)Continuous-time version of the unit-step function of unit amplitude.

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Example 1.8 Rectangular Pulse

Consider the rectangular pulse x(t) shown in Fig. 1.39 (a).Fig. 1.39 (a). This pulse has an amplitude A and duration of 1 second. Express x(t) as a weighted sum of two step functions.

<Sol.><Sol.>

, 0 0.5( )

0, 0.5A t

x tt

1. Rectangular pulse x(t): (1.55)

1 1( )

2 2x t Au t Au t

(1.56)

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Figure 1.39 (p. 44)(a) Rectangular pulse x(t) of amplitude A and duration of 1 s, symmetric about the origin. (b) Representation of x(t) as the difference of two step functions of amplitude A, with one step function shifted to the left by ½ and the other shifted to the right by ½; the two shifted signals are denoted by x1(t) and x2(t), respectively. Note that x(t)

= x1(t) – x2(t).

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Figure 1.40 (p. 45)(a) Series RC circuit with a switch that is closed at time t = 0, thereby energizing the voltage source. (b) Equivalent circuit, using a step function to replace the action of the switch.

1. Initial value: (0) 0v

0( )v V

/( )0( ) 1 ( )t RCv t V e u t (1.57)

2. Final value:

3. Complete solution:

Example 1.9 RC CircuitFind the response v(t) of RC circuit shown in Fig. 1.40 (a).Fig. 1.40 (a).<Sol.><Sol.>

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★ ★ 1.6.6 Impulse Function1.6.6 Impulse Function

Figure 1.42 (p. 46)(a) Evolution of a rectangular pulse of unit area into an impulse of unit strength (i.e., unit impulse). (b) Graphical symbol for unit impulse. (c) Representation of an impulse of strength a that results from allowing the duration Δ of a rectangular pulse of area a to approach zero.

Figure 1.41 (p. 46)Discrete-time form of impulse.


1, 0[ ]0, 0

nnn

(1.58)

Fig. 1.41Fig. 1.41

Figure 1.41 (p. 46)Discrete-time form of impulse.

(t) a(t)

][n

a

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( ) 0 for 0t t (1.59)

( ) 1t dt

(1.60)

Dirac delta function

1. As the duration decreases, the rectangular pulse approximates the impulse more closely.

Fig. 1.42.Fig. 1.42.2. Mathematical relation between impulse and rectangular pulse function:

0( ) lim ( )t x t

(1.61) 1. x(t): even function of t, = duration.

2. x(t): Unit area.Fig. 1.42 (a).Fig. 1.42 (a).

3. (t) is the derivative of u(t):

(1.62)

4. u(t) is the integral of (t):

( ) ( )t

u t d

(1.63)

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<Sol.><Sol.>

1. Voltage across the capacitor:

2. Current flowing through capacitor:

( )( )

dv ti t C

dt 0 0

( )( ) ( )

du ti t CV CV t

dt

Example 1.10 RC Circuit (Continued)For the RC circuit shown in Fig. 1.43 (a),Fig. 1.43 (a), determine the current i (t) that flows through the capacitor for t 0.

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◆ Properties of impulse function:

1. Even function: ( ) ( )t t (1.64) 2. Sifting property:

0 0( ) ( ) ( )x t t t dt x t

(1.65)

3. Time-scaling property:

1( ) ( ), 0at t a

a (1.66)

<p.f.><p.f.>

0

1lim ( ) ( )x at t

a

(1.68)

00

0 00

1( )

IV I t dt

C C

(1.69)

Fig. 1.44Fig. 1.44

1. Rectangular pulse approximation:

0( ) lim ( )at x at

(1.67)

2. Unit area pulse: Fig. 1.44(a).Fig. 1.44(a).

Time scaling: Fig. 1.44(b).Fig. 1.44(b).Area = 1/aRestoring unit area ax(at)

Ex. RLC circuit driven by impulsive source: Fig. 1.45.Fig. 1.45.

For Fig. 1.45 (a)Fig. 1.45 (a), the voltage across the capacitor at time t = 0+ is

)()(lim0

tatax

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Figure 1.44 (p. 48)Steps involved in proving the time-scaling property of the unit impulse. (a) Rectangular pulse xΔ(t) of amplitude 1/Δ and duration Δ, symmetric about the origin. (b) Pulse xΔ(t) compressed by factor a. (c) Amplitude scaling of the compressed pulse, restoring it to unit area.

Figure 1.45 (p. 49)(a) Parallel LRC circuit driven by an impulsive current signal. (b) Series LRC circuit driven by an impulsive voltage signal.

a/ a/

1

1

a

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★ ★ 1.6.7 Derivatives of The Impulse1.6.7 Derivatives of The Impulse

1. Doublet:

(1)

0

1( ) lim ( / 2) ( / 2)t t t

(1.70)

2. Fundamental property of the doublet:

(1) ( ) 0t dt

(1.71)

0

(1)0( ) ( ) ( ) t t

df t t t dt f t

dt

(1.72)

2 (1) (1)(1)

2 0

( / 2) ( / 2)( ) ( ) lim

d t tt t

t dt

(1.73)

3. Second derivative of impulse:

★ ★ 1.6.8 Ramp Function1.6.8 Ramp Function

1. Continuous-time case:

Problem 1.24Problem 1.24

0

2(2)

0 2( ) ( ) ( ) |t t

df t t t dt f t

dt

0

( )0( ) ( ) ( ) |

nn

t tn

df t t t dt f t

dt

, 0( )

0, 0

t tr t

t

(1.74) ( ) ( )r t tu tor (1.75) Fig. 1.46Fig. 1.46

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, 0[ ]

0, 0

n nr n

n

(1.76)

Figure 1.46 (p. 51)Ramp function of unit slope.

or

[ ] [ ]r n nu n (1.77)

Figure 1.47 (p. 52)Discrete-time version of the ramp function.

x[n]

n1 2 3 40123

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Fig. 1.47.Fig. 1.47.

Example 1.11 Parallel CircuitConsider the parallel circuit of Fig. 1-48 (a) involving a dc current source I0 and an initially uncharged capacitor C. The switch across the capacitor is suddenly opened at time t = 0. Determine the current i(t) flowing through the capacitor and the voltage v(t) across it for t 0.<Sol.><Sol.>1. Capacitor current: 0( ) ( )i t I u t

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2. Capacitor voltage:

1( ) ( )

tv t i d

C

0

0

0

0

1( ) ( )

0 for 0

for 1

( )

( )

tv t I u d

Ct

It t

CItu t

CIr t

C

Figure 1.48 (p. 52)(a) Parallel circuit consisting of a current source, switch, and capacitor, the capacitor is initially assumed to be uncharged, and the switch is opened at time t = 0. (b) Equivalent circuit replacing the action of opening the switch with the step function u(t).

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1.7 Systems Viewed as Interconnections of Operations1.7 Systems Viewed as Interconnections of Operations

A system may be viewed as an interconnection of operationsinterconnection of operations that transforms an input signal into an output signal with properties different from those of the input signal.

1. Continuous-time case:


( ) { ( )}y t H x t (1.78)

[ ] { [ ]}y n H x n (1.79) Figure 1.49 (p. 53)Block diagram representation of operator H for (a) continuous time and (b) discrete time.Fig. 1-49 (a) and Fig. 1-49 (a) and

(b).(b).Example 1.12 Moving-average system

Consider a discrete-time system whose output signal y[n] is the average of the three most recent values of the input signal x[n], that is

1[ ] ( [ ] [ 1] [ 2])

3y n x n x n x n

Formulate the operator H for this system; hence, develop a block diagram representation for it.

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21(1 )

3H S S

<Sol.><Sol.> 1. Discrete-time-shift operator Sk: Fig. 1.50Fig. 1.50.

Figure 1.50 (p. 54)Discrete-time-shift operator Sk, operating on the discrete-time signal x[n] to produce x[n – k].

Shifts the input x[n] by k time units to produce an output equal to x[n k].

2. Overall operator HH for the moving-average system:

Fig. 1-51.Fig. 1-51.

Fig. 1-51 (a): cascade form; Fig. 1-51 (b): parallel Fig. 1-51 (a): cascade form; Fig. 1-51 (b): parallel form.form.

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Figure 1.51 (p. 54)Two different (but equivalent) implementations of the moving-average system: (a) cascade form of implementation and (b) parallel form of implementation.

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1.8 Properties of Systems1.8 Properties of Systems

★ ★ 1.8.1 Stability1.8.1 Stability

1. A system is said to be bounded-input, bounded-output (BIBOBIBO) stable if and only if every bounded input results in a bounded output.

2. The operator HH is BIBO stable if the output signal y(t) satisfies the condition

( ) for allyy t M t (1.80)

whenever the input signals x(t) satisfy the condition

( ) for allxx t M t (1.81)

Both Mx and My represent some finite positiv

e number

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Figure 1.52a (p. 56)Dramatic photographs showing the collapse of the Tacoma Narrows suspension bridge on November 7, 1940. (a) Photograph showing the twisting motion of the bridge’s center span just before failure. (b) A few minutes after the first piece of concrete fell, this second photograph shows a 600-ft section of the bridge breaking out of the suspension span and turning upside down as it crashed in Puget Sound, Washington. Note the car in the top right-hand corner of the photograph.

(Courtesy of the Smithsonian Institution.)

One famous example of an unstable One famous example of an unstable system:system:

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Example 1.13 Moving-average system (continued)Show that the moving-average system described in Example 1.12 is BIBO stable.<p.f.><p.f.>

1. Assume that: [ ] for all xx n M n

2. Input-output relation:

1[ ] [ ] [ 1] [ 2]

3y n x n x n x n

1[ ] [ ] [ 1] [ 2]

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[ ] [ 1] [ 2]31

3

x x x

x

y n x n x n x n

x n x n x n

M M M

M

The moving-average The moving-average system is stable.system is stable.

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Example 1.14 Unstable systemConsider a discrete-time system whose input-output relation is defined by

[ ] [ ]ny n r x nwhere r > 1. Show that this system is unstable.

<p.f.><p.f.>

1. Assume that: [ ] for all xx n M n 2. We find that

[ ] [ ] [ ]n ny n r x n r x n ．

With r > 1, the multiplying factor rn diverges for increasing n.

The system is unstable.The system is unstable.

★ ★ 1.8.2 Memory1.8.2 Memory

A system is said to possess memory if its output signal depends on past or future values of the input signal.A system is said to possess memoryless if its output signal depends only on the present values of the input signal.

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Ex.: Resistor1

( ) ( )i t v tR

Memoryless !

Ex.: Inductor1

( ) ( )t

i t v dL

Memory !

Ex.: Moving-average system

1[ ] ( [ ] [ 1] [ 2])

3y n x n x n x n Memory !

Ex.: A system described by the input-output relation2[ ] [ ]y n x n Memoryless !

★ ★ 1.8.3 Causality1.8.3 CausalityA system is said to be causal if its present value of the output signal depends only on the present or past values of the input signal.A system is said to be noncausal if its output signal depends on one or more future values of the input signal.

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1[ ] ( [ ] [ 1] [ 2])

3y n x n x n x n Causal !


1[ ] ( [ 1] [ ] [ 1])

3y n x n x n x n Noncausal !

Causality is required for a systems to be capable of operating in real timereal time.

★ ★ 1.8.4 Invertibility1.8.4 Invertibility

A system is said to be invertible if the input of the system can be recovered from the output.

Figure 1.54 (p. 59)The notion of system invertibility. The second operator H

inv is the inverse of the first operator H. Hence, the input x(t) is passed through the cascade correction of H and H

inv completely unchanged.

1. Continuous-time system: Fig. 1.54Fig. 1.54.

x(t) = input; y(t) = output

H = first system operator;H

inv = second system operator

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2. Output of the second system:

( ) ( ) ( )inv inv invH y t H H x t H H x t 3. Condition for invertible system:

invH H I (1.82) I = identity operator

H inv = inverse operator

Example 1.15 Inverse of System

Consider the time-shift system described by the input-output relation

00( ) ( ) ( )ty t x t t S x t

where the operator S t0 represents a time shift of t0 seconds. Find the inverse of

this system.<Sol.><Sol.>1. Inverse operator S

t 0:

0 0 0 0 0{ ( )} { { ( )}} { ( )}t t t t tS y t S S x t S S x t 2. Invertibility condition:

0 0t tS S I 0tS Time shift of t0

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Example 1.16 Non-Invertible SystemShow that a square-law system described by the input-output relation

2( ) ( )y t x tis not invertible.<p.f.><p.f.> Since the distinct inputs x(t) and x(t) produce the same output y(t).

Accordingly, the square-law system is not invertible.

★ ★ 1.8.5 Time Invariance1.8.5 Time Invariance

A system is said to be time invariance if a time delay or time advance of the input signal leads to an identical time shift in the output signal. A time-invariant system do not change with time.

Figure 1.55 (p.61) The notion of time invariance. (a) Time-shift operator St0 preceding operator H. (b) Time-shift operator St0 following operator H. These two situations are equivalent, provided that H is time invariant.

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1. Continuous-time system:

1 1( ) { ( )}y t H x t2. Input signal x1(t) is shifted in time by t0 seconds:

02 1 0 1( ) ( ) { ( )}tx t x t t S x t S

t 0 = operator of a time shift equal to t0

3. Output of system H:

0

0

2 1 0

1

1

( ) { ( )}

{ { ( )}}

{ ( )}

t

t

y t H x t t

H S x t

HS x t

(1.83)

4. For Fig. 1-55 (b),Fig. 1-55 (b), the output of system H is y1(t t0):

0

0

0

1 0 1

1

1

( ) { ( )}

{ { ( )}}

{ ( )}

t

t

t

y t t S y t

S H x t

S H x t

(1.84)

5. Condition for time-invariant system: 0 0t tHS S H (1.85)

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Example 1.17 Inductor

x1(t) = v(t)

y1(t) = i(t)

The inductor shown in figure is described by the input-output relation:

1 1

1( ) ( )

ty t x d

L

where L is the inductance. Show that the inductor so described is time invariant.<Sol.><Sol.>1. Let x1(t) x1(t t0) Response y2(t) of the inductor to x1(t t0) is

0

1 0 1

1( ) ( )

t ty t t x d

L

2 1 0

1( ) ( )

ty t x t d

L

2. Let y1(t t0) = the original output of the inductor, shifted by t0 seconds:

3. Changing variables: 0' t

(A)

(B)

(A)0

2 1

1( ) ( ') '

t ty t x d

L

Inductor is time invariant.

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Example 1.18 Thermistor

Let R(t) denote the resistance of the thermistor, expressed as a function of time. We may express the input-output relation of the device as

x1(t) = v(t)

y1(t) = i(t)

1 1( ) ( ) / ( )y t x t R tShow that the thermistor so described is time variant.<Sol.><Sol.>1. Let response y2(t) of the thermistor to x1(t t0) is

1 02

( )( )

( )

x t ty t

R t

2. Let y1(t t0) = the original output of the thermistor due to x1(t), shifted by t0

seconds:

1 01 0

0

( )( )

( )

x t ty t t

R t t

3. Since R(t) R(t t0) 1 0 2 0( ) ( ) for 0y t t y t t Time variant!

1 introduction chapter 1.6 elementary signals ★ 1.6.1 exponential signals (1.31) b and a are real...

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