Note 6: Bode Plots @ 2021-03-13 16:54:59-08:00 EECS 16B Designing Information Devices and Systems II Spring 2021 Note 6: Bode Plots Overview Having analyzed our first order filters and gone through a design example in the previous Note to show why filter-design is important, we will now plot their transfer functions H (w) (or frequency responses) using Bode Plots. In the previous Note, we generated tables of jH (w)j, ]H (w) at certain key values of wc, and while this gave some intuition, it didn’t really show what happens at intermediate frequencies. There is immense value in visualizing transfer functions across a wide range of frequencies. Throughout this section, we will use numerical approximations, which will not only prove useful in plotting a filter’s frequency response but also will help us better understand its behavior. 1 Bode Plots When we make Bode plots, we plot the frequency and magnitude on a logarithmic scale, and the angle in either degrees or radians. We use the logarithmic scale because it allows us to break up complex transfer functions into its constituent components. Let’s start by generating Bode Plots for low-pass and high-pass first order filters, which will build our intuition. We will soon see how the analysis of more complex transfer functions can be broken down into parts. 1.1 Low-pass Filter Recall our generalized model of a low-pass filter (perhaps RC or LR): 1 HLP(w) = 1 + jw=wc 6 We plot the magnitude of the frequency response assuming wc = 10 : Note how jHLP(w)j is very close to Linear Approximation 102 Exact jHLP(w)j j 0 ) 10 w ( LP H −2 j 10 10−4 101 102 103 104 105 106 107 108 109 w 1 for w < wc and jHLP(w)j starts dropping off with slope −1 after wc: We can look at 3 distinct regions on the plot: EECS 16B, Spring 2021, Note 6: Bode Plots 1 Note 6: Bode Plots @ 2021-03-13 16:54:59-08:00 • w wc: =) jw=wc ≈ 0: So, HLP(w) ≈ 1 and jHLP(w)j ≈ 1: • w = w : =) H(w) = 1 . So, jH (w)j = p1 : c 1+ j LP 2 wc wc • w wc: =) w=wc 1: Therefore HLP(w) ≈ − j w . So, jHLP(w)j ≈ w : On a log scale, this means 1 that logjHLP(w)j ≈ logwc − logw explaining behavior of dropping off with slope −1. Now let’s plot the phase of HLP(w): ]HLP(w) is very close to 0 for w < 0:1wc and ]HLP(w) is approxi- Linear Approximation 0◦ Exact ]HLP(w) ) w ( LP ◦ H −45 ] −90◦ 101 102 103 104 105 106 107 108 109 w p mately − 2 for w > 10wc: • w 0:1wc =) jw=wc ≈ 0: So, HLP(w) ≈ 1 and ]HLP(w) ≈ 0: 1 ◦ • w = 0:1wc =) HLP(0:1wc) = 1+ j 0:1 and ]HLP(w) ≈ −6 : 1 ◦ • w = wc =) HLP(wc) = 1+ j and ]HLP(w) = −45 : 1 ◦ • w = 10wc =) HLP(10wc) = 1+ j 10 and ]HLP(w) ≈ −84 : 2 ◦ • w 10wc =) w=wc 10: So, HLP(w) ≈ − j · 0 and ]HLP(w) ≈ −90 : We can now better understand the values of the magnitude and phase at 0:1wc, wc, 10wc (as seen in the tables of Note 5). 1.2 High-pass Filter We can similarly analyze our generalized high-pass filter model (CR, RL): jw=wc HHP(w) = 1 + jw=wc 6 we plot the magnitude of the frequency response, again assuming wc = 10 : Here, jHHP(w)j rises with slope 1 for w < wc and jHHP(w)j ≈ 1 after wc: We analyze the plot: • w w ; then w=w 1: Therefore H (w) ≈ j w which implies jH (w)j ≈ w : On a log scale, c c HP wc HP wc this means that logjHHP(w)j ≈ logw − logwc, which explainsthe rising slope of 1: • w = w ; then H(w) = j meaning jH (w)j = p1 c 1+ j HP 2 • w wc; then w=wc 1: Therefore HHP(w) ≈ 1 which implies jHHP(w)j ≈ 1: Now let’s plot the phase of the transfer function HHP(w): 1 Recall that the line y = mx + b has slope m: In this case y = logjHLP(w)j and x = logjwj: 2The magnitude will always be greater than 0; meaning its phase will still be very close to −90◦ EECS 16B, Spring 2021, Note 6: Bode Plots 2 Note 6: Bode Plots @ 2021-03-13 16:54:59-08:00 Linear Approximation 102 Exact jHHP(w)j j ) 100 w ( HP −2 H j 10 10−4 101 102 103 104 105 106 107 108 109 w 90 Linear Approximation Exact ]HHP(w) ) w ( HP 45 H ] 0 101 102 103 104 105 106 107 108 109 w p ]HHP(w) is very close to 2 for w < 0:1wc and ]HHP(w) is approximately 0 for w > 10wc: ◦ • w 0:1wc =) jw=wc ≈ 0: So, HHP(w) ≈ 0 (but slightly positive) and so ]HHP(w) ≈ 90 : j 0:1 ◦ • w = 0:1wc =) HHP(0:1wc) = 1+ j 0:1 and ]HHP(w) ≈ 84 : j ◦ • w = wc =) HHP(wc) = 1+ j and ]HHP(w) = 45 : j 10 ◦ • w = 10wc =) HHP(10wc) = 1+ j 10 and ]HHP(w) ≈ 6 : ◦ • w 10wc =) w=wc 10: So, HHP(w) ≈ 1 and ]HHP(w) ≈ 0 : 2 Second Order Filters We will now consider more complex systems and, in doing so, see the value in appropriate visualizations for transfer function behavior. 2.1 Band-Pass Filters With the knowledge of low-pass filters that block out higher frequencies and high-pass filters that block out lower frequencies, how could we build a filter that lets a specific range of frequncies through? One idea could be to take the output of the low-pass filter and treat it as an input to the high-pass filter. EECS 16B, Spring 2021, Note 6: Bode Plots 3 Note 6: Bode Plots @ 2021-03-13 16:54:59-08:00 CH RL v (t);V center ecenter + vin(t);Vein vout(t);Veout CL − RH Figure 1: A Buffered Band-Pass filter, composed of low-pass and high-pass filter components. Notice the op-amp serving as a unity gain buffer between the two filters. It is introduced to prevent the second circuit from loading the first. The input voltage phasor is Vein at some frequency w. Suppose the two filters have cutoff frequencies wLP and wHP, respectively. Thus, we can write that: Vecenter = HLP(w)Vein Since vcenter(t) is the second filter’s input, the output voltage phasor Veout is: Veout = HHP(w)Vecenter = HHP(w)HLP(w)Vein Thus, the net transfer function HBP(w) is: HBP(w) = HLP(w)HHP(w) More generally, placing filters in series produces a circuit whose transfer function is the product of the indi- vidual transfer functions. From the properties of complex numbers and logarithms, we see that logjz1z2j = logjz1j + logjz2j. So, when plotting jHBP(w)j on a log-log plot, this corresponds to a graphical addition of the individual transfer functions. Note that we are still taking the product of the actual transfer functions, but it resembles a geomtrical sum: jHLP(w)j + jHHP(w)j. We see this idea on display in the examples at the 3 end of this note. Similarly, for ]HBP(w), we can again plot the sum, ]HLP(w) + ]HHP(w). We can compute HBP(w) symbolically as: 1 jwRHCH HBP(w) = HLP(w)HHP(w) = · 1 + jwRLCL 1 + jwRHCH To find the cutoff frequencies of this filter, we can look at the points at which H (w ) = p1 : But based BP c 2 on our approximations from before and from the cuttof frequencies w = 1 and w = 1 , we can LP RLCL HP RHCH approximate jH(w )j ≈ p1 · 1 and jH(w )j ≈ 1 · p1 : This approximation holds best when the cuttof LP 2 HP 2 frequencies are spaced apart. We’ve shown a convenient result! The cutoffs for the band-pass filter are identical to the individual cutoffs 6 4 for the low and high-pass filters. We now plot HBP(w) with wLP = 10 and wHP = 10 to demonstrate the band-pass behavior. 3We must be careful, however, to note that in most of our plots, the x-axis does not correspond to 0, so we can’t simply "stack" the two plots. EECS 16B, Spring 2021, Note 6: Bode Plots 4 Note 6: Bode Plots @ 2021-03-13 16:54:59-08:00 101 jHBP(w)j 100 j ) −1 w 10 ( BP −2 H 10 j 10−3 10−4 101 102 103 104 105 106 107 108 109 w Our band-pass filter uses an op-amp, but what if we cascade our two filters, causing a loading effect? CH RL + + + Vin Vmid CL RH Vout − − − We leave the derivation as an exercise (feel free to post your work on Piazza!), but computing the transfer function yields: jwR C H(w) = H H (1 + jwRLCL)(1 + jwRHCH ) + jwRLCH The loading effect introduces a term of jwRLCH in the denominator. The relative size of this term determines the impact on the circuit. We plot some examples of the band-pass filter with identical low and high cutoff frequencies but different RLCH values to show this loading effect. 100 0 10−4 −3 j −1 10 ) 10 −2 w 10 ( HP H −2 j 10 10−3 101 102 103 104 105 106 107 108 109 w Note how the maximum value of H (w) decreases as RLCH increases.