EE225 FALL 2008 DAVID PARKER

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Abstract—Circuit analysis in the s domain uses simple

algebraic equations and basic DC nodal analysis techniques to predict the circuit behavior. S domain analysis can accurately predict the response of a linear circuit to a variety of different input signals. In particular, the use of a unit impulse source can be used to give one the natural response of the circuit. This can be used to compute the response of the circuit to any source signal.

I. INTRODUCTION series RLC circuit can be used as a bandpass filter or a bandreject filter. The circuit in the title figure, with the

output taken across the resistor is a bandpass filter. This paper examines the above circuit using s domain analysis techniques. The response of the circuit is predicted using these theoretical techniques. These calculated values are then compared to simulated measurements of the circuit (using MapleSim). An actual circuit is breadboarded and some rudimentary frequency response measurements are attempted. The circuit response to impulse, ramp, and step input signals are presented.

The Laplace Transform is the fundamental tool for converting circuits from the time domain to the s domain. Without conversion, circuit analysis requires the use of systems of differential and integral equations. These systems of equations can be tedious and difficult to solve. Once a

Manuscript received Dec. 9, 2008

circuit is converted; the method for solving these s domain

equations requires only simple algebra. Once a circuit has been converted to the s domain via the Laplace transform, the transfer function can be computed. The use of the transfer function facilitates circuit analysis. This will be shown in the theoretical analysis which follows.

II. THEORETICAL ANALYSIS

A. Circuit Description The circuit in the title figure consists of a capacitor, an

inductor and a resistor connected in series with the resistor grounded on one end. The input signal is applied to the high side of the capacitor. The output signal is taken across the resistor. For this analysis, we will be looking at two circuits. These circuits will be denoted as RLC1 and RLC2. The values of the components for these circuits are presented in Table I.

TABLE I

SERIES RLC CIRCUIT VALUES

Circuit R (ohms) L (Henrys) C (Farads)

RLC1 220 0.1 0.1E-6 RLC2 2200 0.1 0.1E-6

An Analysis of a Series RLC Circuit in the s Domain

David F. Parker

A

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B. The Laplace Transform The Laplace transform of a function is given by the

expression

L{ f (t)} = f (t)e! stdt

0

"

# (1)

The Laplace transform of f (t) is also denoted F(s) where

F(s) = L f (t){ } . (2)

The notation used in equation 2 will be employed in the remainder of this paper. In order to analyze a circuit in the s domain, each of the circuit elements must be transformed to the s domain. Table II shows these transformations. Phasor domain transformations are also shown.

TABLE II

CIRCUIT ELEMENT TRANSFORMATIONS

t phasor s

R R R

L j!L sL

C 1

j!C

1

sC

C. The Transfer Function The transfer function is defined as the s domain ratio of the

Laplace transform of the output to the Laplace transform of the input. The transfer function is

H (s) =Y (s)

X(s) (3)

where Y (s) is the Laplace transform of the output signal and X(s) is the transform of the input signal. With this transfer function the output signal (response) of the circuit can be easily determined based on the input signal. If we employ this equation with the series RLC circuit studied, we have

H (s) =R

R + sL +1

sC

(4)

In order to facilitate analysis of this transfer function, we need to put it in standard form. The transfer function in equation (4) is expressed in the standard form where the highest order s term in the denominator is isolated from any coefficients and each term after that is given in decreasing order. For this equation, if we multiply the numerator and the

denominator by s

L, this will do this. The result is

H (s) =

R

L

!"#

$%&s

s2+

R

L

!"#

$%&s +

1

LC

(5)

Substituting the values in RLC1, we have

H (s) =2200s

s2+ 2200s +10

8 (6)

Likewise, for RLC2 we have

H (s) =22000s

s2+ 22000s +10

8 (7)

If we factor the denominator of equations (6) and (7), this

reveals the poles of the transfer function. The poles are defined as the roots of the denominator. For RLC1,

H (s) =2200s

s+1100+9939I( ) s +1100 ! 9939I( ) (8)

and for RLC2, we have

H (s) =22000s

s+15582( ) s + 6417( ) (9)

The poles are indicators of the circuit’s response to an input

signal. If the poles are complex as in RLC1, the circuit will oscillate. If the poles are real as in RLC2, then the circuit decays with an exponential function.

D. The Expression of Vo(s) & Vo(t)-output Based on equation (3), we know that the s domain output of

a circuit can be expressed as Y (s) = H (s)X(s) (10)

If we know the transfer function of a circuit H (s)( ) and we

apply a known input signal X(s)( ) , we can calculate the

output Y (s)( ) . This is also known as Vo(s). If we perform the inverse Laplace transform on this, we have Vo(t)-the output signal in the time domain. The process of returning to the time domain is known as the inverse Laplace transform,

L

!F(s){ } = f (t) .

III. FREQUENCY RESPONSE

The transfer function of a bandpass filter is expressed as

H (s) =!s

s2+ !s +"

0

2 (11)

where ! is the bandwidth (radians/s) and !0

is the center frequency (radians/s). If we examine equation (5) and let

! =R

L, then we can see that !

0=

1

LC. It is clear that

the transfer function of both RLC1 and RLC2 show that they are bandpass filters. !

c1, the lower cutoff frequency

(radians/s) of this filter can be expressed as

!c1= "

#2+

#2

$%&

'()2

+!0

2 . (12)

The cutoff frequency is defined as the frequency where the

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magnitude of the output =1

2Vi. Knowing ! and !

c1, the

upper cutoff frequency can be determined by !

c2=!

c1+ " (13)

Table III

Table III shows the calculated values of !

0,!

c1, and !

c2,

simulated values of !c1

,!c2

,as well as values measured with a breadboard circuit. The values marked with a ‘*’ I was unable to measure with my rudimentary measurement setup.

circuit Calculated values Simulated values Measured values ω0 ωc1 ωc2 ωc1 ωc2 ωc1 ωc2 RLC1 10000 8960 11160 8800 11000 6300 17000 RLC2 10000 3866 25866 3900 25000 * *

Figure 1 and 2 show Bode plots of the two bandpass

circuits. One can see that the frequency response of RLC2 is much more broad and the phase response change is much more gradual than RLC1. The peak in the frequency response plot is the same for both circuits. This agrees with our analysis in that the center frequency is dependent only on the values of the inductor and the capacitor. As the value of R

increases, bandwidth increases as well. Again this agrees

with the equation ! =R

L, as discussed before.

IV. IMPULSE RESPONSE

The Laplace transform of several functions discussed in this paper are presented in Table IV.

TABLE IV

LAPLACE TRANSFORMS

Type f (t)(t > 0!)

F(s)

Impulse ! (t) 1

Ramp t 1

s2

Step u(t) 1

s

The impulse response is the signal that exits a system when

a delta function (unit impulse) is the input. An impulse is a signal composed of all zeros, except a single nonzero point. For our analysis, in order to see the response of the circuit to an impulse, all we need to do is multiply the transfer function of the circuit by 1 (see Table IV and equation (10). Then one performs the inverse Laplace transform to get back to the time domain. In order to facilitate this, we first perform a Partial Fraction Expansion of the transfer function. For RLC1, the result of this operation on equation (8) is

Y (s) =1106.7! + 6.3

!

s +1100 " 9939I+1106.7!" 6.3

!

s +1100 + 9939I (14)

and for RLC2 –equation(9) is

Y (s) =37404.3

(s+15582.)!15404.3

s+6417.4 (15)

The rules for performing the inverse transform of the above

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two forms are given in Table V. TABLE V

TWO USEFUL TRANSFORM PAIRS

Nature of Roots

F(s) f (t)

Distinct real K

s + a Ke

!atu(t)

Distinct Complex

K

s +! " #I+

K*

s +! + #I 2 K e

!"tcos(#t +$)u(t)

Applying the above rules, we have for RLC1

Vo(t) = 2213e

!1100tcos(9939t + 6.3

!

)"# $%u(t) (16)

and for RLC2

Vo(t) = 37404.4e-15582t

-15404.3e-6417t!" #$u(t) (17)

These two equations indicate the impulse response of the circuits. For RLC1, we have what appears to be a damped oscillation. For RLC2, we have a decaying exponential. Figures 3 and 4 show the simulated impulse response.

The above simulations appear to confirm our calculations. Note that the period of oscillations in Figure 3 appears to have the same period as the resonant frequency calculated earlier in the frequency response section of this paper.

V. RAMP RESPONSE

Again, per Table IV and equation (10), we can calculate the circuit response to a ramp input. One simple needs to multiply the transfer function by the s domain representation of a ramp, 1

s2

!"#

$%&

. For RLC1 using equation(8), this gives us

2200s

s+1100+9939I( ) s +1100 ! 9939I( )

1

s2

"#$

%&'

.

For RLC2 using equation(9), this gives us 22000s

s+15582( ) s + 6417( )

1

s2

!"#

$%&

. After Partial Fraction

Expansion (PFE) of the result and use of the rules in Table V for transforming back to the time domain we have for RLC1

Vo(t) = 2.2x10

!5+ 2.2x10

!5e!1100t

cos(9939t +174!

)"# $%u(t)

(18)

and for RLC2

Vo(t) = 2.2x10!4+1.54x10

-4e-15582t ! 3.74x10-4e-6417t"# $%u(t)

(19) Figures 5 shows the simulated ramp input. Figures 6 and 7

show the response to this input. Again, RLC1 shows the characteristic oscillation or ringing at the approximate resonant frequency. RLC2 shows a decaying exponential.

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VI. STEP RESPONSE We can calculate , per Table IV and equation (10), the

circuit response to a step input. One simple needs to multiply the transfer function by the s domain representation of a step, 1

s

!"#

$%&

. For RLC1 using equation (8), this gives us

2200s

s+1100+9939I( ) s +1100 ! 9939I( )

1

s

"#$

%&'

.

For RLC2 using equation (9), this gives us 22000s

s+15582( ) s + 6417( )

1

s

!"#

$%&

. After Partial Fraction

Expansion (PFE) of the result and use of the rules in Table V for transforming back to the time domain we have for RLC1

Vo(t)= .2213e

-1100tcos(9939t+90

!

)!" #$u(t) (20)

and for RLC2

Vo(t) = 2.4e-6417t

-2.4e-15582t!" #$u(t) (21)

Figures 8 shows the simulated step input. Figures 9 and

10 show the response to this input.

VII. CONCLUSIONS

Circuit analysis in the s domain is useful because it allows one to use simple algebra to solve transient response problems. Otherwise, one would have to use complex systems of differential equations. Using the unit impulse source to drive a circuit facilitates analysis of the circuit because this input yields the natural response of the circuit. The unit impulse response of a circuit contains enough information to compute the response of the circuit to any input signal. Measurements made with a small breadboard circuit did not agree very well with simulated or calculated values because the equipment the author used-a $30 signal generator and a NI DAQ with LabView did not have enough accuracy at the higher test frequencies used.