Bode Plots

The zero and pole notation is especially useful for making approximate graphs of the transfer function which are called Bode plots. A Bode plot is a piecewise straight line approximation of the frequency dependence of the transfer function on a log-log scale. It provides an excellent and simple illustration of the frequency response of a circuit. To make a Bode plot of the magnitude versus frequency of the transfer function, on the horizontal axis, we write the log of the angular frequency $\omega$. (Recall $s=j\omega$.) On the vertical axis we write dB, which is defined as $20 log \left\vert\frac{v_o(s)}{v_{in}(s)}\right\vert$.In other words, if $\left\vert\frac{v_o(s)}{v_{in}(s)}\right\vert=100$, then we say the gain is 40dB, and if $\left\vert\frac{v_o(s)}{v_{in}(s)}\right\vert=1000$.then we say the gain is 60dB, etc. Using these axes, the Bode plot is then constructed as an approximate, log-log plot of the transfer function using straight lines in the following way: Starting at low frequencies, and moving toward higher ones, for each zero, we change the slope of the transfer function by +20dB/decade, where a decade is a power of 10 change in frequency. For each pole, we change the slope of the transfer function by -20dB/decade (recall $s=j\omega)$.

Let's first illustrate this with a relatively arbitrary example by constructing a Bode plot of the transfer function in equation 4.8 with $\frac{v_{out}}{v_{in}}$given by  

$$\frac{v_{out}}{v_{in}} =\frac{10^6(s+10)(s+10^2)}{(s+10^3)(s+10^4)(s+10^6)}$$ (73)

So the above expression has zeros of z₁=-10 and z₂=-10², and has poles of p₁=-10³, p₂=-10⁴, and p₃=-10⁶. Next, let's factor the poles and zeros out from each term:

 

$$\frac{v_{out}}{v_{in}}=\frac{10^6*10*10^2}{10^3*10^4*10^6} \... ...2}+1)}{(\frac{s}{10^3}+1) (\frac{s}{10^4}+1)(\frac{s}{10^6}+1)}$$ (74)

To obtain a Bode plot we now start by taking the 20log of the coefficient which is

$20log \left(\frac{10^6*10*10^2}{10^3*10^4*10^6}\right)=-80$We now start plotting at frequency one decade less than the minimum pole or zero, beginning by starting to draw a horizontal line at -80dB from $\omega = 1 rad/sec$ to 10 rad/sec. At $\omega=10 rad/sec$, we encounter a zero, so we increase the slope to 20dB/dec. We continue drawing a straight line with this slope until we reach $\omega= 100 rad/sec$,where we encounter another zero, and thus increase the slope another 20dB/dec to a total slope of 40dB/dec. We now continue along this slope until we reach the first pole which is at $\omega = 10^3 rad/sec$ where we decrease the slope by 20dB/dec to 20dB/dec. We now continue at this slope until we reach the second pole where decrease the slope another 20dB/dec to 0dB/dec. We then continue to draw this horizontal line until we reach the final pole which will give a drop of 20dB/dec. The resulting Bode plot, which is shown in Fig. 4.4.

  

Figure 4.4: Bode Plot Illustration of Equation (4.9)

It is important to notice that the largest value the Bode plot obtained was 0dB. This not an accident. It results because the original expression given by equation (4.8) was characteristic of passive circuits, which are circuits that do not have active gain elements such as transistors or op-amps.

Now, let's obtain a Bode plot for our original circuit of Fig. 4.2. The equation describing frequency-dependent gain of that circuit is given by equation 4.6. Let's factor out the poles to obtain a form similar to that of equation (4.9).  

$$\frac{v_{out}}{v_{in}}= \left(\frac{-R_C\vert\vert R_L}{R_{E... ...2R_{in}C_1(R_L+R_C)C_2}{(sR_{in}C_1+1)(sC_2(R_L+R_C)+1)}\right]$$ (75)

First let's plot the frequency-dependent fractional part, and then add the constant coefficients. Starting with the numerator, we have 2 zeros at $\omega=0$, this gives rise to a line with slope of 40dB/dec which is zero dB at $\omega=1$. At $\omega =\vert p_1\vert$ the slope changes to 20dB/dec, and finally at $\omega = \vert p_2\vert$ the slope decreases by another 20dB/dec to become a horizontal line. Now we add 20log(R_inC₁)(C₂(R_L+R_o)). to obtain the response of the passive part of the transfer function only. Finally, we add the midband gain $20log\frac{R_C\vert\vert R_L}{1/g_m+R_E}$to obtain a graph of the entire equation.

We have just illustrated in detail the mechanics of drawing a Bode plot. However, they can usually be drawn very quickly for the midband to low frequency part of a response with the following approach. Start by finding the highest frequency low-frequency pole, and draw a horizontal line at that point for the midband gain. Next, move to the left (direction of decreasing $\omega)$, and for each pole you encounter increase your slope downward by 20dB/dec, and for each zero you encounter decrease your downward slope by 20dB/dec. In a sense, it is a very quick method of obtaining the same result we described in detail above. The resulting Bode plot is sketched in Fig. 4.5.

  

Figure 4.5: Bode Plot Illustration of Equation (4.10)