measurements using Oscilloscope

Purpose
The purpose of this laboratory is to learn how to use oscilloscope and signal
generator to measure frequency responses of simple RC and RL circuits.
Part I: RC Circuits and its Frequency Response
+
Signal Gen

C

To Scope
Leads

R

_

Consider the above circuit with C=0.1µF and R=1KΩ. Set the output voltage of a
signal generator to 1V (peak-to-peak) in sine wave, that is, VS(t)=cos(ωt) where ω is
the frequency in (radian/second). The impedance expression of the capacitor in the

circuit is XC=1/SC where S{•}={•} can be considered as a derivative operator in
setting up differential equations or S=jω a complex variable of frequency in
calculating frequency responses. Using the impedance notation XC and R, we can
use the Ohm’s law or Voltage divider to find:

VR=+
R or VR=+ VS=+1/SCVS=+1VS

where VR is the voltage across the resistor R and H(s)=

+1

is so called transfer

function. Since VS(t)=cos(ωt) is a sinusoidal signal, the impedance of the capacitor
can be expressed as XC=1/(jωC) in unit of Ω (S= jω). In the circuit,
RC=103(Ω)x10-7(F)=10-4

10−4

H(jω)= +1 = 1+10−4
1

EE 2010 PreLaboratory 8
RC and RL circuits and their frequency responses
measurements using Oscilloscope
The magnitude of a complex number + = √2 + 2 .
Assignment 1: Show the derivation of the magnitude of H(jω) as the follows.
|H(jω)|=

10−4
√1+(10−4 )2

Then
VR=|H(jω)|VS=

10−4
√1+(10−4 )2

VS

Using the relationship: = 2 (Hz=radian/second divided by radian= 1/second).
Considering the VS is 1V (peak-to-peak), calculate the listed values with the
variation of frequencies and fill in the Table 1.
Table 1:
Frequency ω in Frequency Impedance
|H(jω)|
VR (peak-torad/sec
f in Hz
|XC|= 1/ωC (Ω)
peak)
250
1K
5K
10K (ω=1/RC)
20K
100K
1M
Plot the value |H(jω)| (vertical-axis) with respect to the frequency in Hz
(horizontal-axis). (Tip: You can use Matlab to calculate and plot the data or Excel.
The other way is to use Multisim Frequency Response to sweep from 40 Hz to
160KHz.)
1. Explain why it is called a high-pass frequency selective filter.
2. When ω=ωc=1/RC, the frequency is called cutoff (or corner) frequency of the
filter. What is the ratio of VR/VS at the cutoff frequency?
Part II: RL Circuits and its Frequency Response
Consider an RL circuit below with L=100 mH (milli-Henry) and R=1KΩ. Set the
output voltage of signal generator to 1V (peak-to-peak) in sine wave that is,
VS(t)=cos(ωt). The impedance expression of the inductor in the circuit is XL=LS.
2

EE 2010 PreLaboratory 8
RC and RL circuits and their frequency responses
measurements using Oscilloscope
+
Signal Gen

L

To Scope
Leads

R

_
Assignment II: Using Ohm’s law or Voltage divider method, show the derivation
steps to reach the following relationship:

VR=+VS
Since VS(t)=cos(ωt) is a sinusoidal signal, the impedance of the capacitor can be
expressed as XC=jωL (Ω). In the circuit (S= jω), we have
L/R=100×10-3 (H)/103(Ω)=10-4

H(jω)= +R =

1

1+
R

1

= 1+10−4

Finding the expression of magnitude |H(jω)| and considering the VS is 1V (peak-topeak), calculate the listed values with variation of frequencies in the Table 2.
Table 2:
Frequency ω in Frequency Impedance
|H(jω)|
VR (peak-torad/sec
f in Hz
|XL|= ωL (Ω)
peak)
250
1K
5K
10K (ω=R/L)
20K
100K
1M
Plot the value |H(jω)| (vertical-axis) with respect to the frequency in Hz
(horizontal-axis).
1. Explain why it is called a low-pass frequency selective filter.
2. When ω=ωc=R/L, the frequency is called cutoff (or corner) frequency of the
filter. What is the ratio of VR/VS at the cutoff frequency?