Lecture 28
•Review:
• Frequency domain circuit analysis
• Superposition
•Frequency domain system characterization
• Frequency Response
•Related educational modules:
–Section 2.7.4, 2.7.5, 2.8.0
• Note during summary that we will be
changing our mindset, rather than doing
anything fundamentally new
– I will keep re-introducing the same example
circuit, but with minor modifications and
extensions
Frequency domain analysis – review
• The analysis techniques we used for time domain
analysis of resistive networks are applicable to
frequency domain analysis of general circuits
• E.g. KVL, KCL, circuit reduction, nodal analysis, mesh
analysis, Thevenin’s and Norton’s Theorems…
• In general, we simply need to:
• Subsitute impedances for resistances
• Use phasor voltages & currents in place of time domain
voltages & currents
Superposition
• Superposition in frequency domain:
• If multiple signals exist at different frequencies,
superposition is the only valid frequency domain
approach
• Effects of individual sources can be analyzed
independently in the frequency domain
• Summation of individual contributions must be done in
the time domain (unless all contributions have same
frequency)
Example 1 – Superposition
• Determine the steady-state voltage across the inductor in
the circuit below
3W
1W
+
6cos(9t) A
v(t)
-
1
H
3
+
-
4cos(3t+30°) V
• In previous slide, show time domain
superposition. V(t) = v1(t) +v2(t)
– Note that we can determine v1(t) and v2(t)
individually using frequency domain analysis
– We can’t however, superimpose these individually,
for reasons that will be made clear on the next
slide
Example 1 – continued
• The associated frequency domain circuits are shown below:
• Derive the phasors V1 and V2 on the previous
slide
• Point out that they can’t be added directly –
they don’t even apply to the same frequency
domain circuit
– We will superimpose the results in the time
domain
Example 1 – continued again
• Superimpose results in the time domain:
V 1  9 245°
V2 
4
101.6°
10
• On previous slide, V1 has frequency of 9
rad/sec and V2 has frequency of 3 rad/sec
Notes on superposition
• Superposition of sinusoidal signals is extremely
important in circuit analysis!
• In later courses, we will see that (nearly any) signal can
be represented as a superposition of sinusoidal signals,
using Fourier Series and Fourier Transforms
• If we determine the circuit’s response to each of these
sinusoids and superimpose them, we can determine the
circuit’s response to (nearly any) input function
• We will spend the next couple of lectures looking at
circuit responses to multiple sinusoidal inputs
Example 2
• Determine the capacitor voltage, v(t), for the circuit below if
vS1(t) = 3cos(2t+20°) and vS2(t) = 5cos(2t-45°)
• On the previous chart, draw single-source
circuit and point out that VS = vs1+vs2 and
v(t)=v1+v2
– Draw two frequency-domain circuits ; note that
they both have the same appearance
Try a trick to do example 2 easily
• Let’s look at an “arbitrary” input with the same frequency as
our individual inputs:
• Determine input-output relationship (gain,
phase) => point out that this is only good for
2 rad/sec frequency sinusoids
Example 2 – continued
• Now we can use our trick to easily determine the circuit’s
response to both inputs (since they are at the same frequency)
 1

V  V S    45° 
 2

V S 1  3 20°
V S 2  5  45°
Important result
• The steady-state sinusoidal response of a circuit, at a
frequency 0, can be characterized by two values:
• Gain: output-to-input amplitude ratio
• Phase difference: difference between the output and
input phases
• On previous slide, annotate output to show on
block diagram:
– Y = H*U, where H is a complex number that
depends only on frequency
– Circuit “looks like” H, at the frequency of interest
Example 3 – Multiple input frequencies
• Determine the capacitor voltage, v(t), for the circuit below if
vS1(t) = 3cos(2t+30°) and vS2(t) = 5cos(4t+60°)
• Annotate slide to show two frequency-domain
circuits; note that they have different
frequency-domain characteristics
– I don’t want to analyze two circuits, so let’s leave
frequency in the circuit model!
Try a similar trick with example 3
• Let’s characterize the circuit’s gain and phase as a function of
frequency:
Example 3 – continued
• Now we can use our trick to determine the circuit’s response
to both inputs and superimpose the results
V
4
H ( j ) 

V S 4  j 2
vS1(t) = 3cos(2t+30°)
vS2(t) = 5cos(4t+60°)
• On the previous slide, write H, U, and Y as
functions of frequency:
– H(j2), U(j2), Y(j2)
– H(j4), U(j4), Y(j4)
• Note that we analyzed the circuit only once!
Super-important result
• The steady-state sinusoidal response of a circuit is
characterized by the frequency response, H(j)
• Magnitude response: output-to-input amplitude ratio vs.
frequency
• Phase response: difference between the output and input
phases as a function of frequency
Frequency Response
• In the time domain, we characterize systems by the
differential equation relating the input and output
• In the frequency domain, we characterize systems
by their frequency response
• Magnitude response and phase response give the gain
and phase difference relating the input and output
• The frequency of the signal (rather than time) is now our
independent variable