This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHAPTER 14 — FREQUENCY RESPONSE
List of topics for this chapter :
Transfer Function
Bode Plots
Series Resonance
Parallel Resonance Passive Filters
Active Filters Scaling
TRANSFER FUNCTION
Problem 14.1 ' Given the circuit in Figure 14.1 and i(t) = Icos(cot) amps, ﬁnd the transfer function H(0)) = V() / I and sketch the frequency response.
10 Q 20 mH i(t) 0 Figure 14.1 > Carefully DEFINE the problem.
Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem.
To obtain the transfer function, H((o), we need to obtain the frequencydomain equivalent of
the circuit by replacing resistors, inductors, and capacitors with their impedances R, jcoL, and l/jcoC respectively. Then, use any circuit technique to obtain H(03). The frequency response of the circuit can be obtained by plotting the magnitude and phase of the transfer
function as the frequency varies. > Establish a set of ALTERNATIVE solutions and determine the one that promises the
greatest likelihood of success.
After transforming the circuit from the time domain to the frequency domain, we can use
nodal analysis, mesh analysis, or basic circuit analysis to ﬁnd the transfer function. Let's
examine the frequency—domain equivalent circuit in order to make the best choice. 243 Transforming the circuit to the frequency domain yields V1 10 Q V0
I ‘l E jw(0.02) Q It seems obvious that nodal analysis will yield two equations with three unknowns (1, V1, and
V0). These equations can be manipulated to eliminate V1 in order to ﬁnd H(w). > ATTEMPT a problem solution.
Using nodal analysis, AVl—O Vl—V Atnodel: 1+ 10 + 10 =0 Vo — V1 V0 — 0
+ 7— : 0
10 JCO(0.02) At node 2 : Using the equation for node 1, ﬁnd V1 in terms of V0 and I . 101=2V1—V0 . _Vo+101 V —l’ 51
I“ 2 ‘2+ Simplify the equation for node 2.
jco(Vo — V1) + 500Vo = 0 (jw+500)Vo—jcoV1=0 Now, substitute the equation for V1 into the simpliﬁed equation for node 2. ‘
(jco+ 500)V0 — jco(0.5V0 + 51) = 0' I ‘ jco/2+500 '1000+jm Therefore,
jml 0 H =————
(w) 1000+jco > EVALUATE the solution and check for accuracy.
A check of our solution can be done using basic circuit analysis. Find the output voltage as
the current through the inductor multiplied by the impedance of the inductor, i.e., V = I Z. . 244 Label the necessary variables for this technique. 10 Q
lo
I 0 jco(0.02) Q I _,__£__I__1_9_I
° " 10+10+ jco(0.02) “ 20+ jw(0.02) V =ZI_WI_ jwlo ° ‘ 20+ jco(0.02) _1000+jco
XL_ jcolO
I "1000+jw Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not,
then return to “ALTERNATIVE solutions” and continue through the process again. . This problem has been solved satisfactorily. jcolO H =—————
(0’) 1000+ jco The freg uency response is 10' (1000, 7.071) Magnitude (unitless) 103 10
Frequency (radians/second) 245 Problem 14.2 Given the circuit in Figure 14.2 and Vin (t) = Vin cos(0)t) volts, ﬁnd the .
transfer function H(0)) = Vout/ Vin and sketch the frequency response. 10kg luF Figure 14.2 Transform the circuit to the frequency domain. 10 k!) 410%) Q Clearly,
Vout = jw(0.01)I
d I Vin
an = .
106
104 J +jo)(0.01)
(0
Thus,
'106
Vin =[104  J m + jco(0.01))1
So, _ jw(0.01)1 _ JO)
V _ ' 6 J a ( 1‘”)
m 1 4 _ ° . 6 ___~___
[0 m +_]0)(001)I 10 +JCO 0)
Therefore,
H(O)) = —Jmi0T
106 + {03— ——)
0) 246 The freg uency response is O. 1 .0
(O .0
on .0
‘1 .0
a 0.5 Magnitude (unitless) 10“ 105 10° 10 10
Frequency (radians/second) Problem 14.3 Given the circuit in Figure 14.3 and i(t) = Iin cos((ot) amps, ﬁnd the
transfer function H(c0),= Vout / I in and sketch the frequency response. Figure 14.3 The fre uenc res onse is 10000 9000 8000 7000 6000 5000 4000 Magnitude (unitless) 3000
2000 1000 0
. 1o3 10‘ 105
Frequency (radians/second) 247 for the transfer function 104 Hm) : to 106
1 ' —~—
+Jiloz 0) ) WW BODE PLOTS
Problem 14.4 Sketch the Bode plots, both magnitude and phase, given the following
transfer function in the sdomain.
90 s +1 3 +10
ms) : ( x x )
s (s + 3)(s + 30) First, we need to modify the transfer function so that it is in a form that is easy to plot. S S S . S
(90)(s+1)(s+10)_ (“Damiiﬂ i6“) (10%“ 1'6“) s(s+3)(s+30) “ S S = S S
(3)(30)(s)(—3+1 ﬁn) (S)(§+1 5+1) Begin with a plot showing the magnitude curve _of each term in the transfer function. H(S) = 20 log10(10) 20
20 log10(j(0 + 20 log10(1/a)) 20 loglo(j((o/10)+1) 20 log10(1/[j((2)/30)+1]) 20 logm(1/[j(c0/3)+1]) 248 Now, combine, or add, the curves to acquire the composite magnitude (dB) plot of the transfer
. function. Note that the dashed curve shows the approximation to the actual curve. 90° +*1
(i(co/10)+1) 30 100 300 (l/[j(0)/30)+1])
‘_J (1/[j(oo/3)+1]) —90° Finally, the composite phase angle plot for the transfer function can be drawn. 249 Problem 145 Given the following Bode plot, determine the value of the transfer function,
H(s), represented by the Bode plot. 0.1 1 10 100 1000 From this ﬁgure we have the following H(s), determined by looking at each break point and
realizing that the slope everywhere is incremented by 20 dB per decade. Please note that the one break point between 60 = 1 and 10 is estimated to be equal to 3.
l 5 +1)
s ~_
10
H(s) = i K s s s
. 1 W _~ m
(s + )[3 + ljtoo + lX1000 + I) where K is given by 2010g10 (K) = 40 —> K = ﬂ Problem 14,6 . Given the following Bode plot, determine the value of the transfer function,
H(s), represented by the Bode plot. ~20 dB —40 dB —60 dB 250 Decomposing the Bode plot into its basic elements we get, 20 dB 0 dB
0.01 I 0.1 20 log10(1/10)
~20 dB 1/((s/100) + 1) *__J 1/((s/0.3) + 1) l/((s/10) + 1) Therefore,
+ 1
ms) = i W_§_____
(—s—+lIi+l in)
0.3 10 100
SERIES RESONANCE Problem 14.7 Given the circuit in Figure 14.4, ﬁnd:
(a) the resonant frequency and the halfpower frequencies,
(b) the bandwidth and the quality factor, (0) the amplitudes of Vc (t) at coo , (01, and (02. 1 uF 10 mH 10 cos(c0t) Figure 14.4 251 (a) (b) (c) 1 The resonant frequency of a series RLC circuit is 030 = Vt; .
LC So,
a) 1 1 104 d/
= —'—— = ——— = ra s
0 J LC 4102 x106
(00 = 10 krad/s
. . . . _ R
The halfpower frequency of a series RLC crrcult 1s 031,2 = + E +
So,
4 10 +[ 10 )2+[ 1
“"3 ‘ 2x10‘2 2x102 102 x106
031,2 = —T 500+ J (500)2 +108
031,2 = T 500 + 10,012 rad/s
or
(01 = 9.512 krad/s
£02 2 10.512 krad/s
The bandwidth of a series RLC circuit is B = 032 —— col .
So,
B =10.512—9.512
B = l krad/ s
. . . . c00
The quality factor of a serles RLC c1rcu1t 15 Q = B“.
So,
Q _ y:
— 103
Q =19
The amplitudes of vC (t) at 030, 031 ,and (02
—j106/m s2 jco10'2 g2
1040° 10 Q 252 J R 1 (ET  ' At 0) = (00 = 10 krad/s, the inductor has a value of j104 ><10‘2 = leO ohms and the
capacitor has a value of  j106 /104 = j100'ohms. Then, w=mml
but
10
‘1: 10+j100—j100 :1 amp
So,
VC = j100 volts
and vC (t) = 100 cos(10,000t — 90°) volts This gives an amplitude of 100 V at 0) = 030 = 10 krad/ 5. At 0) = (1)1 = 9.512 krad/s, the inductor has a value of j9,512 x 0.01 = j95.12 ohms
and the capacitor has a value of  j106/9,512 = j105. 13 ohms. Then, vC = j105.13I
but
10 10
I=10+j95.12—j105.13 :10—j10.01 amps
So,
VC = (105.13290°)(0.7068245.03°) = 74.32—44.97° volts
and vC (t) = 74.3oos(9,512t — 44.97°) volts This gives an amplitude of 74.3 V at (0 = (01 = 9.512 krad/s. At co = (02 = 10.512 krad/s, the inductor has a value of j10,512 x 0.01: j105.12 ohms
and the capacitor has a value of — j106 /10,512 = j95.13 ohms. Then, vC = j95.13I
but
10 10
I: 10+j105.12—j95.13 =10+j9.99 amps
So,
VC = (95.132  90°)(0.70752  44.97°) = 67.32 134.97° volts
and vC (t) = 67.3cos(10,512t—134.97°) volts This gives an amplitude of 67.3 V at 0) = (02 = 10.512 krad/s. 253 Note that the output voltage for this bandpass ﬁlter is the voltage across the resistor. It
can be shown that vout (t) = 10cos(10,000t) V at a) = 030 = 10 krad/s
vout (t) = 7.068cos(9,512+ 45.03) V at to = 001 = 9.512 krad/s
vout (t) = 7.075 cos(10,512t — 44.97°) V at to = a)2 = 10.512 krad/s The amplitude at the halfpower frequencies is 1/ ‘5 times the maximum amplitude at
the center frequency. In this case, 1
$00) = 7.071 where the calculated amplitudes of 7.068 volts and 7.075 volts are quite close to the
expected halfpower value of 7.071 volts. Problem 14.8 Given the circuit in Figure 14.5, ﬁnd the value of L so that we have a Q of
100. Also, ﬁnd (00, £01, 032, and B. 10 Q L
20 cos(0)t) 1 uF
Figure 14.5
L = 1 H
030 =1krad/s w1=995radls m2=1005radls B=10rad/s PARALLEL RESONANCE Problem 14.9 Given the circuit in Figure 14.6 and I = 240° amps, ﬁnd
(a) 030, Q, and B,
(b) col and 032,
(0) power dissipated at 030 ,0)] , and 032. 254 Figure 14.6
. . . 1
(a) The resonant frequency of a parallel RLC circuit 15 (00 = E.
So,
1 104
0) = m :
° duo2x106)
030 = 1 krad/ s
. . . R
The quality factor of a parallel RLC c1rcu1t IS Q = 0) L .
0
So,
Q _ L
’ 104 ><10'2
Q = 100
. . . 030
The bandwidth of a parallel RLC Circuit 15 B = So, I
B _ a
7 100
B = 100 rad / s
(b) Because this is a high Q circuit, the halfpower frequencies can be written as
__ B
031,2 = (00 + '2“
So,
4 _ 100
(01,2 = 10 + ‘2‘
0)],2 = 104 T 50
or 031 = 9.95 krad/ s
032 = 10.05 krad/s 255 (c) Find the power dissipated at 000 = 1 krad/s. So, 1
1
=§<2)2<104) 1
P = 20 kwatts 1
Since all of the current ﬂows through the resistor at resonance, P = I i 2 R . Since col and (02 correspond to the halfpower points, the power dissipated at col and
032 is 10 kwatts. Problem 14.10 Given the circuit in Figure 14.7, ﬁnd the resonant frequency. 109 10 mH Figure 14.7 . 1 l
l Begin by ﬁnding the parallel equivalent of the series resistor and inductor elements. The parallel
equivalent is given by So,
1 1 1 ________ = _+
10+ jun/100 Req jer
10—j03/100 1 1 m +
100+ (co/100)2 R jX 9‘! 5‘1 Thus,
jco/100 _ 1
100+(co/100)2 “ leq
_ 100+(m/100)2 X“! (5/100 256 At resonance, . (0:030 and Xeq =XC
X ___1_____1___E
Where C _ (DC m 0310‘6 — a)
BLEDLE:
Thus’ (no 100 ‘ (00
or
104 +‘°—‘2’—1o6
100 ‘ cog =108 ~106 =9.9><107
(00 = 9.95 krad/s or
coo E 10 krad/s «mm . PASSIVE FILTERS Problem 14.11 What type of ﬁlter is represented by the circuit in Figure 14.8? What is the
cutoff frequency, or what are the corner frequencies? lOkQ 10kQ Figure 14.8 In the frequency domain, the circuit is 10k!) 257 Find the transfer function Hm) : Vent . V. in Using nodal analySiS, Vout _ Vin + Vout Vout = 0 104 —j105/03+ 104 Simplifying,
, 0)
Vout _ Vin + J—l—avout + Vout = 0
[2 + 'gjv — V out — in
Hence,
H = ——
(w) 2 + jco/l 0
This transfer function looks like a typical transfer function for a lowpass ﬁlter
1
1 + ijC l
5 Vin and as to —> oo Vout = 0 , We can look at a value for 0.7071
=( 2 jive To ﬁnd the cutoff frequency, ﬁnd the value of 0) when IH(co)I = 0.3535 . Since the voltage starts at Vout = = 0.3535 lvin out V lH(co)l=——1—O)2~=0.3535
4+ﬁ
(Dz
8 =4+1()6
(02 = 400
a): 20 rad/s This lowpass ﬁlter has a cutoff frequency of (0C = 20 rad/ s or fC = 10/ 7t Hz . 258 Problem 14.12 [14. 43] Determine the range of frequencies that will be passed by a
series RLC bandpass ﬁlter with R = 10 Q, L = 25 mH, and C = 0.4 MP. Find the quality
factor. 1 1
coo =——=————————=1o krad/s
x/LC ,/(25 x10‘3)(0.4 x10'6) R 10
B = — = ‘_—‘ = 4
L 0.025 00 rad/S
Thus,
Q_ 9132
‘ B “ o.
Q=_5_
This is a high Q circuit so we can use
9.8
(01 =030—B/2=10—0.2=9.8 krad/s or fl 22::15597kHz
10.2
(02 :030+B/2=10+0.2=10.2 krad/s or f2 =~§=L6234kﬂz Therefore,
1.5597 kHz < f < 1.6234 kHz Pro’blem 14.13 What type of ﬁlter is represented by the circuit in Figure 14.9? What is the
cutoff frequency, or what are the corner frequencies? 10k§2 Vin O 10 k9 Figure 14.9 This highpass ﬁlter has a cutoff frequency of (0C = 26.55 rad/ s or fC = 4.226 Hz . 259 ACTIVE FILTERS Problem 14.14 What type of ﬁlter is represented by the circuit in Figure 14.10? What is the
cutoff frequency, or what are the corner frequencies? 100 kg Figure 14.10 In the frequenCy domain, the circuit is Find the transfer function V011!
H(03) = V. m 260 Using nodal analysis, . Va_Vin+Va—Vout+Va—Vout_0
105 105 j106/co ‘ where Va = Vb = 0. Simplifying,
‘Vm _ Vout —ji_(_)Vout : 0
(1+ .39.}, _
J 1 0 out — 1n
Hence,
H 0) = ————
( ) 1+ jm/ 10
This transfer function looks like a typical transfer function for a lowpass ﬁlter
1
1+ ijC This lowpass ﬁlter has a cutoff frequency of (DC = 10 rad/ s or fc = 1.5915 Hz. Problem 14.15 [14. 55] Design the ﬁlter in Figure 14.11 to meet the following requirements : »
(a) It must attenuate a signal at 2 kHz by 3 dB compared with its value at 10 MHz. (b) It must provide a steadystate output of V0 (t) = 10 sin(27t x 108 t + 180°) volts
for an input vi (t) = 4 sin(27t x108 t) volts. Figure 14.11 This is a highpass ﬁlter with
. f0 = 2 kHz, (x)C = 27th =1/(RC) 261 or
1 1 RC 2 27:1; 2 471x103 Clearly, the capacitor becomes a short circuit at high frequencies. Hence, the high frequency gain
is Rf 10
———=—— or R = 2.5R
R 4 f IfweletR:10kQ,then Rf=25kQ,andC== =7.958nF. 40001: x 104 SCALING Problem 14.16 [14.63] For the circuit in Figure 14.12, (a) draw the new circuit after it has been scaled by Km = 200 and Kf = 104.
(b) obtain the Thevenin equivalent impedance at terminals a—b of the scaled circuit at co = 104 rad/s. Figure 14.12 (a) R' = KmR = (200)(2) = 400 :2 K L (200)(1)
I: ——"—1— = — = 20
L K 104 262 We now have a new circuit, b
_ (b) Insert a 1 amp source at the terminals a—b.
a V1 SL V2
HM
1A 0 0 0.51K
b —:
At node 1 : At node 2 :
. V V —V V —V V
1: 1 1 2 1 2 = __2_
1/(sC) + sL sL +0‘SIX R
But, Ix = sCVl.
So, the nodal equations become
V1 —V2 V1 —V2 V2
1—sCV1 + SL sL +0.55CV1=E
Solving for V],
V __ sL+R
‘ ' $2LC+O.5sCR+l
Z _ 1a _ ,__sIg_R_.__
Th " 1 " 52LC+0.SSCR+1
'104 20 '3
“03:10:, Zn: . 4 2 a (J )(_6x10 )fr4400
(J10 ) (20x10 )(0.25x10 )+0.5(]10 )(0.25x10'6)(400)+1
z — 9991299 — 600 '200
‘ Th ‘ 0.5+j0.5 “ “J
. zTh = 632.5418.43°§2 263 Problem 14.17 Given the circuit in Figure 14.13, ﬁnd the values necessary to scale this .
circuit, increasing the comer frequency to 100 rad/s. Use a 1 uF capacitor. 291. Figure 14.13 To scale the circuit in Figure 14.13 from 0) = 1/ 4 rad/s to 03’ = 100 rad/s using a 1 HF capacitor,
the feedback resistor and the in ut resistor must be 10 k9. ‘ 264 ...
View
Full Document
 Spring '10
 Chi
 RLC, Nyquist plot

Click to edit the document details