Sketch the Bode plots for the following transfer function:
$$ H(s) = \frac{10s(s+20)}{(s+1)(s^2+60s+400)} $$
Rewrite the expression in a recognizable form for zeros and poles:
We recognize that the transfer function contains a zero at the origin, simple pole, simple zero and a quadratic/complex pole. We begin by rewriting the transfer function in a more standard form: $$ H(s) = \frac{(10s)(20)(1+\frac{s}{20})}{(1+\frac{s}{1})400(\frac{400+60s+s^2}{400})} $$ $$ \qquad \; = \frac{(20)(10)s(1+\frac{s}{20})}{400(1+\frac{s}{1})\Big[1+\frac{3s}{20}+\Big( \frac{s}{20} \Big)^2\Big]} $$
$$ H(s) = \frac{0.5s(1+\frac{s}{20})}{(1+\frac{s}{1})\Big[1+\frac{3s}{20}+\Big( \frac{s}{20} \Big)^2\Big]} \qquad,(Eqn\;1)$$
Determing magnitude via logarithms
If we substitute "jw" for "s" in equation #1 we get: $$ H(j\omega) = \frac{0.5j\omega(1+0.05j\omega)}{(1+j\omega)(1+0.15j\omega-0.0025\omega^2)} \qquad,(Eqn\;2)$$ Recall that the gain/magnitude of a transfer function is defined as: $$ H_{dB} = 20 \log_{10}|\mathbb{H}(j\omega)| $$ If we apply this expression to equation #2 and use the properties of logarithms we get the following expression:
$$ H_{dB} = 20\log_{10}0.5 + 20\log_{10}|j\omega|+20\log_{10}|1+0.05j\omega|-20\log_{10}|1+j\omega|$$ $$\qquad \qquad \qquad \qquad \qquad \qquad -20\log_{10}|1-0.0025\omega^2+j0.15\omega| $$ $$ H_{dB} = -6.02+20\log_{10}\omega+20\log_{10}\sqrt{1+0.0025\omega^2}-20\log_{10}\sqrt{1+\omega^2}$$ $$\qquad \qquad \qquad \qquad \qquad \qquad -20\log_{10}\sqrt{(1-0.0025\omega^2)^2+0.0225\omega^2} $$
$$ H_{dB} = -6.02+20\log_{10}\omega+10\log_{10}(1+0.0025\omega^2)-10\log_{10}(1+\omega^2)$$ $$\qquad \qquad \qquad \qquad \qquad \qquad -10\log_{10}\Big[(1-0.0025\omega^2)^2+0.0225\omega^2\Big] $$
Sketch the magnitude plot by examining equation #1
$$ H(s) = \frac{0.5s(1+\frac{s}{20})}{(1+\frac{s}{1})\Big[1+\frac{3s}{20}+\Big( \frac{s}{20} \Big)^2\Big]} \qquad,(Eqn\;1)$$ By comparing equation #1 to the forms we derived for the Zero at origin, Simple Zero, Simple Pole and quadratic pole, we see that we have the following corner frequencies: $$ \omega_0 = -1, +20, -20$$ ...where the "+" and "-" are simply a means of designating a zero (positive sign) and a pole (negative sign). We sketch the magnitude plot by examining each of the transfer function factors in order of increasing corner frequencies.
First and Second Factors (constant and zero at origin):
In equation #1 we see that the first factor is the constant 0.5. Recalling the section on Bode plots and constant factors, we know that the magnitude for the constant factor in our transfer function is: $$ H_{db} = 20\log_{10}0.5 = -6.02\;dB $$ This means that our starting magnitude is -6.02 dB. However, remember that we don't start our plot at 0 rad/sec along the frequency axis due to the fact that it is logarithmic. Rather, we will begin at 0.1 rad/sec as it is sufficiently close to the origin to give us a reasonably accurate plot over a wide range of frequencies. Our second factor is the Zero at origin: $$ s $$ We know that a single zero at the origin begins at -20 dB and has a straight-line magnitude plot with a slope of +20 db/decade. However, we must combine this with the magnitude of the constant term (-6.02 dB). This gives us the following starting magnitude: $$ Starting\;magnitude = -26 \; dB $$ $$ slope = +20\frac{dB}{dec} \qquad,(0.1 \leq \omega \leq 1) $$ Therefore, we begin our magnitude plot at -26 dB and draw a straight line with a slope of +20 dB/decade until we hit our first corner frequency of 1 rad/sec.
Third factor (simple pole):
The corner frequency of 1 rad/sec is associated with the Simple Pole: $$ \frac{1}{1+\frac{s}{1}} $$ The straight-line magnitude plot of a simple pole has a slope of -20dB/decade. However, this is combined with the previous slope of our 1st and 2nd factors (which is +20 db/dec) We therefore obtain a net slope of 0 db/dec. Proceed to plot a straight horizontal line with a slope of 0 dB/decade from 1 rad/sec until we hit the next corner frequency of 20 rad/sec. $$ slope = 0\frac{dB}{dec} \qquad,(1 \leq \omega \leq 20)$$
Fourth and fifth factors (simple zero and quadratic pole):
The corner frequency of 20 rad/sec is associated with the Simple Zero and quadratic pole: $$ 1+\frac{s}{20} \qquad,\qquad \frac{1}{1+\frac{3s}{20}+\Big(\frac{s}{20}\Big)^2} $$ The straight-line magnitude plot of a simple zero has a slope of +20 dB/decade. Additionally, the straight-line magnitude plot of a quadratic pole has a slope of -40 db/dec. Since these two factors have the same corner frequency, we can add their slopes together to obtain a net slope of -20 db/decade over the same range. We also combine this with the net slope of our previous factors (which is 0 dB/dec) and arrive at a slope of -20 db/dec from 20 rad/sec onwards. $$ slope = -20\frac{dB}{dec} \qquad,(20 \lt \omega)$$
Deviation at 20 rad/sec:
If you recall our discussion of Bode plots for quadratic poles and quadratic zeros of the transfer function, we mentioned how the damping coefficient (zeta) can be used to give more precision to the Bode magnitude plot. Specifically, at the corner frequency of our simple pole, the deviation between the straight-line magnitude plot and the actual plot can be defined as: $$ H_{deviation \; in \; dB} = -20\log_{10}(2\zeta) $$ Compare our quadratic pole: $$ \frac{1}{1+\frac{3s}{20}+\Big(\frac{s}{20}\Big)^2} $$ ...to the standard form for a quadratic pole: $$ \frac{1}{1+2\zeta\Big(\frac{s}{\omega_0}\Big)+\Big( \frac{s}{\omega_0} \Big)^2} $$ By equating terms, we see that for our pole, the following expression holds true: $$ 2\zeta = 3 \qquad,(where\;\zeta=damping\;coefficient) $$ We can now calculate the deviation at the corner frequency of 20 rad/sec: $$ H_{deviation \; @ \; \omega=20} = -20\log_{10}(2\zeta) $$ $$ \qquad \qquad \qquad \; = -20\log_{10}3 $$ $$ \qquad \qquad \qquad \; = -9.5 \; dB $$ We can represent this difference between the approximate and actual plot by drawing a vertical line at 20 rad/sec that goes downward 9.5 dB from the straight-line approximation. A smooth-curve line could then be drawn to better approximate the magnitude plot.
Straight-line approximation for Bode magnitude plot:
After examining the five factors of the transfer function in the manner described above, (and the deviation at the 20 rad/sec corner frequency) the final straight-line approximation of the Bode magnitude plot appears as follows
Exact Bode Magnitude Plot:
Below we see the exact Bode magnitude plot (achieved via software).:
Determine phase using concept of phasors
Recall equation #2: $$ H(j\omega) = \frac{0.5j\omega(1+0.05j\omega)}{(1+j\omega)(1+0.15j\omega-0.0025\omega^2)} \qquad,(Eqn\;2)$$ Also, recall that we can determine the phase angle of any complex phasor quantity by recognizing that: $$ \phi = \tan^{-1} \Big( \frac{imaginary\;part}{real\;part} \Big) $$ Applying this to equation #2 gives us:
$$ \phi = 90^{\circ}+\tan^{-1}(0.05\omega)-\tan^{-1}(\omega)-\tan^{-1}\Big( \frac{0.15\omega}{1-0.0025\omega^2} \Big) \qquad,(Eqn\;3)$$
For the above equation, don't forget about the signs of the real part of the quadratic pole as they apply to reference angles from 180 degrees.
Sketch the phase plot
Simple Pole
Once again, when sketching the phase plot, it is important to remember that for factors of the transfer function not at the origin, the slopes of those factors affect the overall plot over a range of frequency values starting at one decade before to one decade after their respective corner frequencies. Consider the corner frequency of 1 rad/sec associated with the Simple Pole of our transfer function: $$ \frac{1}{1+\frac{s}{1}} $$
- For corner frequency of 1 rad/s:
- $$ 1 \; decade \; before = 0.1 \frac{rad}{s} $$
- $$ 1 \; decade \; after = 10 \frac{rad}{s} $$
A single Simple Pole has a straight-line phase plot with a slope of -45 deg/decade. Therefore, the simple pole in our transfer function imparts a slope of: $$ -45\frac{deg}{dec} \qquad,over \; the \; range\;(0.1 \leq \omega \leq 10) $$
Simple Zero
Now we look at the corner frequency of 20 rad/sec associated with the Simple Zero of our transfer function: $$ 1+\frac{s}{20} $$
- For corner frequency of 20 rad/s:
- $$ 1 \; decade \; before = 0.2 \frac{rad}{s} $$
- $$ 1 \; decade \; after = 200 \frac{rad}{s} $$
A single Simple Zero has a straight-line phase plot with a slope of +45 deg/decade. Therefore, the simple zero in our transfer function imparts a slope of: $$ +45\frac{deg}{dec} \qquad,over \; the \; range\;(0.2 \leq \omega \leq 200 ) $$
Quadratic Pole
The corner frequency of 20 rad/sec is also associated with the quadratic pole of our transfer function: $$ \frac{1}{1+\frac{3s}{20}+\Big(\frac{s}{20}\Big)^2} $$
- Once again, for corner frequency of 20 rad/s:
- $$ 1 \; decade \; before = 0.2 \frac{rad}{s} $$
- $$ 1 \; decade \; after = 200 \frac{rad}{s} $$
A single quadratic pole has a straight-line phase plot with a slope of -90 deg/decade. Therefore, the quadratic pole in our transfer function imparts a slope of: $$ -90\frac{deg}{dec} \qquad,over \; the \; range\;(0.2 \leq \omega \leq 200 ) $$
Constant term and zero at origin
The constant factor of our transfer function is: $$ 0.5 $$ As explained in the section on Bode plots and constant factors, the constant term imparts a phase angle of zero degrees. Additionally, we have the Zero at origin: $$ s $$ The zero at the origin has a constant straight-line phase plot of 90 deg. Once again, you may be tempted to draw a horizontal line at 90 degrees from w=0 up till our first frequency of interest (w=0.1). Remember however that the horizontal axis is logarithmic and as such we never include w=0 on the plot. Instead we use 0.1 rad/sec as our starting point and begin our phase plot at 90 degrees: $$ 90^{\circ} = starting\;phase $$ Remember that this is an approximation. Notice that if you plug 0.1 rad/sec into our phase equation (equation #3) you will get about 83 degrees. If you need more accuracy in this range you could start the plot at 90 degrees at a lower frequency and draw a smooth fit line.
The final straight-line approximation (with annotations) is shown below:
Straight-line approximation for Bode phase plot:
Notice how the imparted slope values for the individual factors have ranges that overlap. Within these overlapping frequency ranges, the slopes of those factors are combined.
Exact Bode Phase Plot:
Below we see the exact Bode phase plot (achieved via software):
In the next page we will see how we can use software (ie: Maple and GNU Octave) to construct Bode plots
Continue on to Bode Plot (software)...