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H ( s ) x ( t ) y ( t ) 1 1 1 c c c x ( t ) = A cos( t + - - PowerPoint PPT Presentation

Jan 25, 2023 146 likes •322 views

Analog Filters Overview To Do List Discuss time-domain versus frequency domain characteristics Ideal Filters Overshoot First-Order Filters Settling time Active & Passive Filters Show examples of filters applied to

slide-1
SLIDE 1

Introduction to Filters H(s)

x(t) y(t)

x(t) = A cos(ωt + φ) yss(t) = A|H(jω)| cos (ωt + φ + ∠H(jω))

  • In general, H(jω) will vary with ω
  • Filters attenuate ranges of frequencies
  • Used in many applications

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Analog Filters Overview

  • Ideal Filters
  • First-Order Filters
  • Active & Passive Filters
  • Second-Order Filters
  • Resonance
  • RLC Filters
  • Impedance & Frequency Scaling

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Ideal Filters

1 Lowpass 1 Highpass 1 Bandpass 1 Bandstop 1 Notch

ω ω ω ω ω ωc ωc ωc ωc1 ωc1 ωc2 ωc2

  • There are five ideal filters
  • Lowpass filters pass low frequencies: ω < ωc
  • Highpass filters pass high frequencies: ω > ωc
  • Bandpass filters pass a range of frequencies: ωc1 < ω < ωc2
  • Bandpass filters pass two ranges: ω < ωc1 and ω > ωc2
  • Notch filters pass all frequencies except ω ∼

= ωc

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To Do List

  • Discuss time-domain versus frequency domain characteristics

– Overshoot – Settling time – Show examples of filters applied to signals

  • Discuss how to view time-domain characteristics

– Explain why impulse response is only appropriate for low-pass and bandpass filters – Show step response of all filters

  • Use problem from Winter 2002 final as an example
  • Go over other popular second-order active filters discussed in

Franco’s text

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SLIDE 2

Example 1: Workspace

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Ideal Filters Comments

1 Lowpass 1 Highpass 1 Bandpass 1 Bandstop 1 Notch

ω ω ω ω ω ωc ωc ωc ωc1 ωc1 ωc2 ωc2

  • Phase is not shown
  • ωc is called the cutoff frequency
  • Generally, the ideal phase is 0◦ for all frequencies
  • Can not build ideal filters in practice
  • Real filters appear as rounded versions of ideal filters
  • Most LTI systems can be thought of as non-ideal filters

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Example 1: Bode Plot

10

−2

10

−1

10 10

1

10

2

−40 −30 −20 −10 |H(jω)| (dB) Passive Lowpass RC Filter 10

−2

10

−1

10 10

1

10

2

−80 −60 −40 −20 ∠ H(jω) (degrees) Frequency (rad/s)

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Example 1: Passive RC Filter

R C vs(t) vo(t)

  • +

H(s) = Vo(s) Vs(s) = 1 RC

  • 1

s +

1 RC

  • 1. What is the DC (ω = 0) response of this filter?
  • 2. What is limω→∞ |H(jω)|?
  • 3. What type of filter to you think this is?
  • 4. Plot |H(jω)| and ∠H(jω) for

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SLIDE 3

Example 2: Cutoff Frequency

R C vs(t) vo(t)

  • +

Find the cutoff frequency for the circuit shown above.

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Example 1: MATLAB Code

w = logspace(-2,2,1000); H = 1./(j*w + 1); subplot(2,1,1); h = semilogx(w,20*log10(abs(H))); set(h,’LineWidth’,1.5); ylabel(’|H(j\omega)| (dB)’); title(’Passive Lowpass RC Filter’); set(gca,’Box’,’Off’); grid on; set(gca,’YLim’,[-45 5]); set(gca,’XLim’,[min(w) max(w)]) subplot(2,1,2); h = semilogx(w,angle(H)*180/pi); set(h,’LineWidth’,1.5); ylabel(’\angle H(j\omega) (degrees)’); set(gca,’Box’,’Off’); grid on; set(gca,’YLim’,[-95 5]); set(gca,’XLim’,[min(w) max(w)]) xlabel(’Frequency (rad/s)’); Portland State University ECE 222 Analog Filters

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Example 2: Workspace

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Cutoff Frequency Defined The cutoff frequency is defined as the value of ω such that 1 2 = |H(jω)|2 maxω |H(jω)|2 = |H(jω)|2 H2

max

where Hmax ≡ max

ω

|H(jω)|

  • Lowpass and highpass filters have a single cutoff frequency
  • Bandpass and bandstop filters have two cutoff frequencies
  • Notch filters have a single notch frequency (not formally defined)

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SLIDE 4

Frequency Selective Impedance

sL R sC I(s) I(s) I(s) 1 V(s)

  • +

V(s)

  • +

V(s)

  • +
  • We can think of passive elements as frequency-selective

impedance sources

  • At low frequencies: s = jω → 0

– R is constant – L approximates a short circuit – C approximates an open circuit

  • This is the same as doing a DC analysis

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Bode Plots & Cutoff Frequency |H(jωc)|2 H2

max

= 1 2 |H(jωc)| Hmax =

  • 1

2 20 log10 |H(jωc)| − 20 log10 Hmax = 20 log10

  • 1

2 HdB(ωc) − HdB,max = −3.0103 dB HdB,max − HdB(ωc) = 3.0103 dB

  • Cutoff frequency always occurs at ≈ 3 dB below Hmax
  • This is also known as the 3 dB frequency

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Frequency Selective Impedance Continued

sL R sC I(s) I(s) I(s) 1 V(s)

  • +

V(s)

  • +

V(s)

  • +
  • At high frequencies: s = jω → ∞

– R is constant – L approximates an open circuit – C approximates a short circuit

  • Inductors are rarely used directly because they are difficult to build

compactly (in silicon)

  • In ECE 321 you will learn how to use op amps and capacitors to

simulate inductors

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Analog Filters Overview

1 Lowpass 1 Highpass 1 Bandpass 1 Bandstop 1 Notch

ω ω ω ω ω ωc ωc ωc ωc1 ωc1 ωc2 ωc2

  • We will initially discuss first and second order filters
  • These are poor filters
  • Poor filters are adequate for many applications
  • These are the building blocks for very good filters
  • Passive filters consist only of passive elements: R, L, and C
  • Active filters also include active elements such as op amps or

transistors

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SLIDE 5

Example 3: First Order Highpass

vs(t) vo(t)

  • +

vo(t)

  • +

vs(t) R1 R2 C R C

Find the transfer functions for the first-order highpass filters shown above.

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First Order Lowpass

R C vs(t) vo(t)

  • +

vo(t)

  • +

vs(t) RL R1 C R2

We derived the transfer functions for these circuits earlier. HP (s) =

1 RC

s +

1 RC

HA(s) = −

1 R1C

s +

1 R2C

H(s) = kωc s + ωc

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Example 3: Workspace

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First Order Lowpass Bode Plot

10

−2

10

−1

10 10

1

10

2

−40 −30 −20 −10 |H(jω)| (dB) Lowpass First−Order Filter 10

−2

10

−1

10 10

1

10

2

−80 −60 −40 −20 ∠ H(jω) (degrees) Frequency (rad/s)

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SLIDE 6

Cascading Transfer Functions G(s)

x(t)

H(s)

y(t)

F(s)

x(t) y(t)

  • In general, the transfer function of cascaded passive filters is not

the product: F(s) = G(s)H(s)

  • For active filters, you can cascade the filters if the output of each

system is connected to the output of an op amp

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Example 3: Bode Plot

10

−2

10

−1

10 10

1

10

2

−40 −30 −20 −10 |H(jω)| (dB) Highpass First−Order Filter 10

−2

10

−1

10 10

1

10

2

20 40 60 80 ∠ H(jω) (degrees) Frequency (rad/s)

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Example 4: Cascaded Circuits

RL CL vs(t) vo(t)

  • +

CH RH Lowpass Highpass

Calculate the transfer function for the cascade of passive filters shown

  • above. Generate the bode plots for the product of the transfer

function of each stage and for the transfer function of the circuit. Use CL = 1 µF, RL = 1 kΩ, CH = 1 µF, RH = 10 kΩ.

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Example 3: MATLAB Code

w = logspace(-2,2,5000); H = j*w./(j*w + 1); subplot(2,1,1); h = semilogx(w,20*log10(abs(H))); set(h,’LineWidth’,1.5); ylabel(’|H(j\omega)| (dB)’); title(’Highpass First-Order Filter’); set(gca,’Box’,’Off’); grid on; set(gca,’YLim’,[-45 5]); set(gca,’XLim’,[10^-2 10^2]) subplot(2,1,2); h = semilogx(w,angle(H)*180/pi); set(h,’LineWidth’,1.5); ylabel(’\angle H(j\omega) (degrees)’); set(gca,’Box’,’Off’); grid on; set(gca,’YLim’,[-5 95]); set(gca,’XLim’,[10^-2 10^2]) xlabel(’Frequency (rad/s)’); Portland State University ECE 222 Analog Filters

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SLIDE 7

Example 4: MATLAB Code

N = 10000; wn = 1; wh = 1e3; wl = 10e3; wlh = 1e3; w = logspace(1,6,N)’; sys1 = tf([wl 0],[1 (wl+wh+(wl*wh)/wlh) (wl*wh)]); [m,p] = bode(sys1,w); mag1 = zeros(size(w)); phs1 = zeros(size(w)); mag1(:) = m(1,1,:); phs1(:) = p(1,1,:); sys2 = tf([wl 0],[1 (wh+wl) wh*wl]); [m,p] = bode(sys2,w); mag2 = zeros(size(w)); phs2 = zeros(size(w)); mag2(:) = m(1,1,:); phs2(:) = p(1,1,:); Portland State University ECE 222 Analog Filters

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Example 4: Solution You should be able to show H(s) = sωL s2 + ( ωHωL

ωHL + ωL + ωH)s + ωLωH

HLP (s) = ωL s + ωL HHP (s) = s s + ωH HLP (s)HHP (s) = ωLs (s + ωL)(s + ωH) = sωL s2 + (ωL + ωH)s + ωLωH where ωL ≡ 1 RLCL ωH ≡ 1 RHCH ωHL ≡ 1 RLCH

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Example 4: MATLAB Code Continued

subplot(2,1,1); h = semilogx(w,20*log10(mag1),’b’,w,20*log10(mag2),’r:’); set(h,’LineWidth’,1.5); grid on; ylabel(’Mag (dB)’); title(’Cascade Passive Example’); axis([10^1 10^6 -40 5]); subplot(2,1,2); h = semilogx(w,phs1,’b’,w,phs2,’r:’); set(h,’LineWidth’,1.5); grid on; xlabel(’Frequency (rad/sec)’); ylabel(’Phase (deg)’); axis([10^1 10^6 -95 95]); legend(’H(s)’,’H_L(s) H_H(s)’); Portland State University ECE 222 Analog Filters

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Example 4: Bode Plot

10

1

10

2

10

3

10

4

10

5

10

6

−40 −30 −20 −10 Mag (dB) Cascade Passive Example 10

1

10

2

10

3

10

4

10

5

10

6

−50 50 Frequency (rad/sec) Phase (deg) H(s) HL(s) HH(s)

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SLIDE 8

Example 5: MATLAB Code

N = 10000; wn = 1; A = 10; wh = 1e3; wl = 10e3; w = logspace(1,6,N)’; sys = tf(A*[wl 0],[1 (wl+wh) (wl*wh)]); [m,p] = bode(sys,w); mag = zeros(size(w)); phs = zeros(size(w)); mag(:) = m(1,1,:); phs(:) = p(1,1,:); Portland State University ECE 222 Analog Filters

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Example 5: Cascaded Circuits

Lowpass Highpass vs(t) R1 CL R2 vo(t)

  • +

R4 CH R3

H(s) = HL(s)HH(s) =

  • −R2

R1

1 R2CL

s +

1 R2CL

−R4 R3 s s +

1 R3CH

  • =

R2R4 R1R3 ωLs (s + ωL)(s + ωH) = k ωLs (s + ωL)(s + ωH) Generate the Bode plot for k = 10, ωL = 10 k rad/s, and ωH = 1 k rad/s.

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Example 5: MATLAB Code Continued

subplot(2,1,1); h = semilogx(w,20*log10(mag),’b’,w,20*log10(mag),’r--’); set(h,’LineWidth’,1.5); grid on; ylabel(’Mag (dB)’); title(’Cascade Active Example’); axis([10^1 10^6 -25 25]); subplot(2,1,2); h = semilogx(w,phs,’b’,w,phs,’r:’); set(h,’LineWidth’,1.5); grid on; xlabel(’Frequency (rad/sec)’); ylabel(’Phase (deg)’); axis([10^1 10^6 -95 95]); legend(’H(s)’,’H_L(s) H_H(s)’); Portland State University ECE 222 Analog Filters

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Example 5: Bode Plot

10

1

10

2

10

3

10

4

10

5

10

6

−20 −10 10 20 Mag (dB) Cascade Active Example 10

1

10

2

10

3

10

4

10

5

10

6

−50 50 Frequency (rad/sec) Phase (deg) H(s) HL(s) HH(s)

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SLIDE 9

Second-Order Filters: HR(s)

10

−2

10

−1

10 10

1

10

2

−60 −40 −20 Mag (dB) Series RLC HR(s) 10

−2

10

−1

10 10

1

10

2

−50 50 Frequency (rad/sec) Phase (deg) Q = 0.1 Q = 0.5 Q = 0.707 Q = 1 Q = 2 Q = 10 Q = 100

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Second-Order Passive Filters

R L C v(t) vR(t)

  • +

vL(t)

  • +

vC(t)

  • +

vLC(t)

  • +

H(s) = N(s)ω2

n

s2 + ωn

Q s + ω2 n

= N(s)ω2

n

s2 + R

Ls + 1 LC

  • Series RLC circuits are used in many types of filters
  • Six possibilities for the output voltage
  • Each is of the same form

– Undamped Natural Frequency: ωn =

1 √ LC

– Quality Factor: Q = L

Rωn = 1 R

  • L

C Portland State University ECE 222 Analog Filters

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Second-Order Filters: HR(s) MATLAB Code

N = 1000; wn = 1; Q = [0.1 0.5 sqrt(1/2) 1 2 10 100]; mag = []; phs = []; w = logspace(-2,2,N); for cnt = 1:length(Q), q = Q(cnt); sys = tf([wn/q 0],[1 wn/q wn^2]); [m, p] = bode(sys,w); mag = [mag reshape(m,N,1)]; phs = [phs reshape(p,N,1)]; end; Portland State University ECE 222 Analog Filters

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Second-Order Passive Filters Continued

R L C v(t) vR(t)

  • +

vL(t)

  • +

vC(t)

  • +

vLC(t)

  • +

HR(s) ≡ VR(s) V (s) =

R Ls

s2 + R

Ls + 1 LC

=

ωn Q s

s2 + ωn

Q s + ω2 n

HL(s) ≡ VL(s) V (s) = s2 s2 + R

Ls + 1 LC

= s2 s2 + ωn

Q s + ω2 n

HC(s) ≡ VC(s) V (s) =

1 LC

s2 + R

Ls + 1 LC

= ω2

n

s2 + ωn

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SLIDE 10

Second-Order Filters: HC(s)

10

−2

10

−1

10 10

1

10

2

−80 −60 −40 −20 20 40 Mag (dB) Series RLC HC(s) 10

−2

10

−1

10 10

1

10

2

−150 −100 −50 Frequency (rad/sec) Phase (deg) Q = 0.1 Q = 0.5 Q = 0.707 Q = 1 Q = 2 Q = 10 Q = 100

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Second-Order Filters: HR(s) MATLAB Code Continued

subplot(2,1,1) h = semilogx(w,20*log10(mag)); set(h,’LineWidth’,1.5); grid on; ylabel(’Mag (dB)’); title(’Series RLC H_R(s)’); axis([10^-2 10^2 -60 5]); subplot(2,1,2); h = semilogx(w,phs); set(h,’LineWidth’,1.5); grid on; xlabel(’Frequency (rad/sec)’); ylabel(’Phase (deg)’); axis([10^-2 10^2 -95 95]); legend(’Q = 0.1’, ’Q = 0.5’, ’Q = 0.707’, ’Q = 1’,... ’Q = 2’, ’Q = 10’, ’Q = 100’); Portland State University ECE 222 Analog Filters

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Second-Order Passive Filters Continued

R L C v(t) vR(t)

  • +

vL(t)

  • +

vC(t)

  • +

vLC(t)

  • +

HRL(s) ≡ VR(s) + VL(s) V (s) = HR(s) + HL(s) = s2 + ωn

Q s

s2 + ωn

Q s + ω2 n

HLC(s) ≡ VL(s) + VC(s) V (s) = HL(s) + HC(s) = s2 + ω2

n

s2 + ωn

Q s + ω2 n

HCR(s) ≡ VC(s) + VR(s) V (s) = HC(s) + HR(s) =

ωn Q s + ω2 n

s2 + ωn

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Second-Order Filters: HL(s)

10

−2

10

−1

10 10

1

10

2

−80 −60 −40 −20 20 40 Mag (dB) Series RLC HL(s) 10

−2

10

−1

10 10

1

10

2

50 100 150 Frequency (rad/sec) Phase (deg) Q = 0.1 Q = 0.5 Q = 0.707 Q = 1 Q = 2 Q = 10 Q = 100

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SLIDE 11

Second-Order Filters: HCR(s)

10

−2

10

−1

10 10

1

10

2

−80 −60 −40 −20 20 40 Mag (dB) Series RLC HCR(s) 10

−2

10

−1

10 10

1

10

2

−150 −100 −50 Frequency (rad/sec) Phase (deg) Q = 0.1 Q = 0.5 Q = 0.707 Q = 1 Q = 2 Q = 10 Q = 100

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Second-Order Filters: HRL(s)

10

−2

10

−1

10 10

1

10

2

−80 −60 −40 −20 20 40 Mag (dB) Series RLC HRL(s) 10

−2

10

−1

10 10

1

10

2

50 100 150 Frequency (rad/sec) Phase (deg) Q = 0.1 Q = 0.5 Q = 0.707 Q = 1 Q = 2 Q = 10 Q = 100

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Second-Order Passive Filters Summary

R L C v(t) vR(t)

  • +

vL(t)

  • +

vC(t)

  • +

vLC(t)

  • +
  • Series RLC circuits can be used to build 4 good second-order

filters – vR(t): Bandpass – vL(t): Highpass – vC(t): Lowpass – vLC(t): Notch

  • Each of these is defined by two parameters: Q and ωn
  • This leaves one degree of freedom to select R, L, and C

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Second-Order Filters: HLC(s)

10

−1

10 10

1

−60 −40 −20 Mag (dB) Series RLC HLC(s) 10

−1

10 10

1

−50 50 Frequency (rad/sec) Phase (deg) Q = 100 Q = 10 Q = 2 Q = 1 Q = 0.707 Q = 0.5 Q = 0.1

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SLIDE 12

2nd Order Filter Impulse Response

  • Recall that if Q ≤ 0.5, the roots are real
  • As Q → ∞, all filters pass only ωn

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Filter Impulse Response

  • You should also be familiar with the impulse response of filters
  • In the time domain, filtering effectively takes a weighted average
  • f the input signal
  • Consider the following plots and how they weight the input signal

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Impulse Response: Second Order Lowpass

0.5 1 1.5 2 2.5 3 3.5 4 −6 −4 −2 2 4 6 h(t) 2nd Order Lowpass Filter Impulse Response Time (sec) Q = 0.25 Q = 0.5 Q = 0.707 Q = 1 Q = 1.414 Q = 5 Q = 10

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Impulse Response: First Order Lowpass

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 5 10 15 20 25 h(t) 1st Order Lowpass Filter Impulse Response Time (sec) ωc = 1 Hz ωc = 2 Hz ωc = 3 Hz ωc = 4 Hz ωc = 5 Hz

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slide-13
SLIDE 13

Revisiting Q Consider the bandpass filter HR(s) =

ωn Q s

s2 + ωn

Q s + ω2 n

You should be able to show that the cutoff frequencies are at ωL = ωn

  • 1 +

1 4Q2 − 1 2Q

  • ωH = ωn
  • 1 +

1 4Q2 + 1 2Q

  • ωn = √ωLωH
  • On a log scale, ωn appears halfway between ωL and ωH
  • Bandwidth is defined as BW = ωH − ωL
  • In this case, Q = ωn

BW

  • This is the definition of Q for higher-order bandpass filters

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Impulse Response: Second Order Highpass

0.5 1 1.5 2 2.5 3 3.5 4 −10 −8 −6 −4 −2 2 4 6 h(t) 2nd Order Highpass Filter Impulse Response ωn = 2 π Time (sec) Q = 0.25 Q = 0.5 Q = 0.707 Q = 1 Q = 1.414 Q = 5 Q = 10

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Impedance Scaling

  • In the process of designing a filter it is often useful to use filter

prototypes with simple element values

  • For example: first order lowpass filter with unity DC gain and

unity cutoff frequency

  • Could use R = 1 Ω and C = 1 F
  • There are two types of scaling

– Impedance or magnitude scaling – Frequency scaling

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Impulse Response: Second Order Bandpass

0.5 1 1.5 2 2.5 3 3.5 4 −1 1 2 3 4 5 h(t) 2nd Order Bandpass Filter Impulse Response Time (sec) Q = 0.25 Q = 0.5 Q = 0.707 Q = 1 Q = 1.414 Q = 5 Q = 10

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SLIDE 14

Example 6: Scaling Use the prototype for a first-order RC highpass filter to design a high-pass filter with a cutoff frequency of 1 k rad/s. Constrain your design so that it uses a 1 µF capacitor.

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Magnitude & Frequency Scaling Goal: Increase the impedance at all frequencies by km Old Impedance New Scaled Impedance Equivalent Value R kmR R′ = kmR Ls kmLs L′ = kmL

1 Cs km Cs

C′ =

1 km C

Goal: Shift impedance at frequency ω to frequency kfω Old Impedance New Scaled Impedance Equivalent Value R R R′ = R Ls L s

kf

L′ =

L kf 1 Cs kf Cs

C′ =

1 kf C Portland State University ECE 222 Analog Filters

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Example 6: Workspace

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Magnitude & Frequency Scaling Continued Combine both of these into a single operation: Old Value New Value R R′ = kmR L L′ = km

kf L

C C′ =

1 kmkf C

kf = ω′ ω km = Z′ Z

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SLIDE 15

Example 8: Scaling

2 F 5 F vs(t) vo(t)

  • +

v1(t)

10 Ω 20 Ω

Suppose that you wish to replace the 2 F capacitor with a 4 µF capacitor and you wish to increase the cutoff frequency by a factor of 10,000.

  • 1. What is the frequency scaling coefficient?
  • 2. What is the impedance magnitude scaling coefficient?
  • 3. Draw the new circuit with the new resistor and capacitor values

below.

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Example 7: Scaling Use scaling to design an RLC lowpass filter with an undamped natural frequency of 10 krad/s and Q = 1. Constrain your design so that it uses a 10 µF capacitor. Hint: recall that Q = 1

R

  • L

C . Portland State University ECE 222 Analog Filters

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Example 8: Workspace

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Example 7: Workspace

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SLIDE 16

Analog Filters Summary

  • There are many types of filters
  • We discussed several types

– Passive & Active – First-order – Second-order – Lowpass & Highpass – Bandpass & Bandstop – Notch

  • Additional filters were covered in lab & the text
  • For a given H(s), there are many implementations & tradeoffs

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