Bode plots and resonance

Two things that only make sense across a range of frequencies rather than at one: the Bode plot, which draws a transfer function's gain and phase as frequency sweeps, and resonance, which is the one frequency where a circuit's reactance vanishes.

11.1Do Bode plots

A Bode plot shows the gain and the phase of a transfer function against frequency, on a logarithmic axis.

Symbulator 9 draws it for you, in the Plot card below the results. Solve the circuit first — the plot is of an answer, so there has to be one — then set Plot type to Bode plot of a variable (FD). The two time boxes relabel themselves as frequencies.

e,1,0,1
r1,1,2,1000
c1,2,0,1e-6

• Variable to plot: v2
• Start frequency (Hz) and End frequency (Hz): 10 and 100000
• Points: 300

Both curves appear together, magnitude in decibels above and phase in degrees below. For this low-pass RC that is the rolloff you would expect: flat at low frequency, falling 20 dB per decade after the corner, with the phase heading for −90°.

When you have H(s) itself

The textbook often hands you the transfer function and no circuit — which is exactly what the calculator's tool took. Version 9 takes it too: set Plot type to Bode plot of a transfer function H(s) (FD), and the variable box becomes Transfer function H(s). Type the function there; the Circuit Description is ignored for this plot type, so it works on an empty page.

• Write the function in terms of s. A textbook function of jω becomes one of s by writing s wherever jω appears — a Bode plot sweeps the imaginary axis, where s is jω.
• The expression takes the same shorthand a circuit value does — ^ for powers, implied multiplication, 1'k — and it must be numeric apart from s. A leftover symbol is refused by name.
• The frequency boxes are in hertz. The calculator examples in this lesson give their ranges in rad/s; divide by 2π for the same sweep in Hz (0.1 to 300 rad/s is about 0.016 to 48 Hz).

11.2Resonance

The resonant frequency is the one at which a circuit's reactance vanishes — where the impedance the source sees becomes purely real.

There is no dedicated tool, and none is needed: solve the circuit in AC with w in the ω — angular frequency box, then put the condition to the Solve card.

Two things make it work, and both are easy to leave out:

• *Tick real solutions only.* Solving im(ze) = 0 over the complex field returns answers written in terms of re(w) and im(w) rather than w — useless. Declaring the unknown real is what makes it solvable.
• Add w > 0 as a condition. The algebra gives ± the frequency, and only one of the two is a frequency.

AS7's Example 14.7

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Find the resonance frequency ω₀ and the bandwidth frequencies ω₁ and ω₂.

Solution

Circuit Description

e,1,0,20
r,1,2,2
l,2,3,1'm
c,3,0,.4'u

Put w in the ω box and solve in AC. Then, in the Solve card:

Equation(s) to solve in terms of the results

im(ze) = 0

Unknown(s) to solve for

w

Conditions

w>0

with real solutions only ticked. It gives ω₀ = 50000 rad/s.

The bandwidth frequencies are the half-power points, where the impedance magnitude is √2 times its resistance. Same circuit, different equation:

Equation(s) to solve in terms of the results

abs(ze) = sqrt(2)*2

With the condition w<50000 that gives ω₁ = 49010 rad/s, and with w>50000 it gives ω₂ = 51010 rad/s.

All three are correct.

AS7's Example 14.8

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Find the resonance frequency and the bandwidth frequencies.

Solution

Circuit Description

e,1,0,10
r,1,0,8'k
l,1,0,.2'm
c,1,0,8'u

Trying the same equation as before does not go well here. The peak of resonance is so sharp — a needle — that a numerical search struggles to find its way to the solution. A different condition converges immediately: ask instead for the frequency at which the source's current has no imaginary part.

Equation(s) to solve in terms of the results

im(ie) = 0

with w>0 and real solutions only, giving ω₀ = 25000 rad/s.

Version 9 solves this symbolically rather than by searching, so the sharpness of the peak costs it nothing — but the alternative equation is worth knowing anyway, because it says something true about the circuit: at resonance the source sees a purely real load, so its current is in phase with its voltage.

AS7's Example 14.9

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Determine the resonant frequency ω₀ of the circuit.

Solution

Circuit Description

j,0,1,1
c,1,0,.1
r1,1,0,10
l,1,2,2
r2,2,0,2

This time, look for the frequency at which the imaginary part of the voltage at node 1 vanishes.

Equation(s) to solve in terms of the results

im(v1) = 0

with w>0 and real solutions only.

ω₀ is 2 rad/s. Correct.