This preview shows page 1. Sign up to view the full content.
Unformatted text preview: driven series RLC circuit can serve as a lowpass ﬁlter. i Frequency L x(t ) +
− Time vL(t ) = Li(t ) VL(s) = sLI (s) R vR(t ) = Ri(t ) VR(s) = RI (s) C Z
+
1t
y(t ) vC (t ) =
i(!)!
C0
− EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly VC (s) = 1
I (s)
sC 26 Circuit Examples for Second Order Filters
A driven series RLC circuit can serve as a lowpass ﬁlter. i Frequency L x(t ) +
− Time vL(t ) = Li(t ) VL(s) = sLI (s) R vR(t ) = Ri(t ) VR(s) = RI (s) Z
+
1t
y(t ) vC (t ) =
i(!)!
C0
− C The factors sL, R, and
resistor, and capacitor. 1
sC VC (s) = 1
I (s)
sC are the complex impedences of the inductor, EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 27 Transfer Function of Second Order Filter i Time L x(t ) +
− R
C Frequency vL(t ) = Li(t ) VL(s) = sLI (s) vR(t ) = Ri(t )
Z
+
1t
y(t ) vC (t ) =
i(!)!
C0
− VR(s) = RI (s)
VC (s) = 1
I (s)
sC In the time domain
x(t) = vL(t) + vR(t) + vC (t); y (t) = vC (t) In the frequency domain (with zero initial conditions)
X (s) = VL(s) + VR(s) + VC (s);
EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly Y (s) = VC (s)
28 X (s) =
Y (s) = 1
sL + R +
sC
1
I (s)
sC I (s) and the transfer function H (s) = Y (s)/X (s)
H (s) =
=
= 1
sC I (s)
1
sL + R + sC I (s)
1
sC
1
sL + R + sC s2 1/(LC )
+ (R/L)s + 1/(LC ) This is a second order lowpass ﬁlter, with DC gain 1 (0 dB).
EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 29 The corner frequency ωc (assuming the underdamped case) is
ωc =
1/(LC ). The total impedence of the series RLC circuit is
ZT (s) = sL + R + 1/(sC )
and the impedence of the output element (the capacitor) is
ZO (s) = 1/(sC )
The transfer function is a voltage divider
H (s) = ZO (s)/ZT (s) = 1
sC 1
sL + R + sC which we compute as if the complex impedences were resisitors.
EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 30 For convenience, we’ll let C = 1, L = 1, and R = √ 2: I (s)
1/s
√
s + 2 + 1/s
1
√
=
s2 + 2s + 1 H (s) = s
√
2 X (s) +
− 1/s +
Y (s)
− Bode Plot 20 logMagnitude (dB) ω))
10 (H ( j 20 ! 10 1 −40 dB/decade 0 −3 dB !10 × !20
!30
!40
!1
10 1 10 ω 0 ∠H ( (deg))
Phase j ω !
× 0 10 !50 !100 −1 !150 !200
!1
10 0 10
Frequency (rad/s) ω 1 10 EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 31 What if we deﬁne the output to be the voltage drop across the inductor?
I (s)
s
√
2 X (s) +
− +
Y (s)
− H (s) = s
√
s + 2 + 1/s = s2
√
s2 + 2s + 1 20 log10 (H ( jω))
Magnitude (dB) 1/s
!
1 Bode Plot
20
10
0 × −3 dB !20
!30
!40
!1
10 −1 0 10 ω 1 10 200 !
∠H ((deg) )
Phase j ω ◦
◦
× +40 dB/decade !10 150 100 50 0
!1
10 EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 0 10
Frequency (rad/s) ω 1 10 32 What if we dene the output to be the voltage drop across the resistor?
I (s)
s
√
2 X (s) +
− 1/s √
2
√
H (s) =
s + 2 + 1/s
√
2s
√
=
2 + 2s + 1
s +
Y (s)
− Bode Plot 20 logMagnitude (dB) ω))
10 (H ( j 20 ! 0 dB 10 1 0 !10 × !20
!30 × ! −1 0 10 ω 1 10 100 ∠Phase jω)
H ( (deg) ◦ −20 dB/decade +20 dB/decade !40
!1
10 50 0 !50 !100
!1
10 0 10
Frequency (rad/s) ω EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 1 10 33 Conclusion
The frequency response of a system with a transfer function (system
function) H (s) is H (j ω ), provided the j ω axis is in the region of convergence
for H (s).
The pole/zero placement determines the frequency response:
• H (j ω ) increases near a pole
• H (j ω ) decreases near a zero
Many polezero patterns have been well studied, such as the Butterworth
lowpass.
Simple approximations accurately characterize many systems. EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 34...
View
Full
Document
This note was uploaded on 09/28/2013 for the course EE 108b taught by Professor Mukamel during the Fall '13 term at Singapore Stanford Partnership.
 Fall '13
 Mukamel
 Frequency, Signal Processing

Click to edit the document details