For this potential divider circuit:
RA
Vin
Vout
(input)
RB
(output)
a) Express Vout in terms of Vin, RA and RB
b) If Vin is a dc voltage of 6V, what is Vout?
c) If Vin is a voltage source which varies with
time according to the equation
Vin(t) = 20 sin (2π × 50 t), what is Vout?
d) If RA = RB, express the ratio Vout / Vin in
decibels.
• This problem is designed to show that potential
dividers behave the same with dc and ac signals, and
the concept of gain expressed in decibels.
• (a) The output voltage is given by the usual formula:
⎛ RB ⎞
Vout = Vin × ⎜
⎟
⎝ RA + RB ⎠
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• (b) With Vin = 6V, the output is obviously
⎛ RB ⎞
Vout = 6 × ⎜
⎟
⎝ RA + RB ⎠ , that is, we just
RB
multiply Vin by the ratio RA + RB .
• (c) If Vin(t) = 20 sin (2π × 50 t), we do the same thing:
⎛ RB ⎞
Vout (t ) = 20 sin ( 2π × 50t) × ⎜
⎟
⎝ RA + RB ⎠
It doesn’t matter that Vin now varies with time; at
each point in time it’s still just multiplied by the factor
RB
RA + RB .
• (d) Clearly the ratio Vout / Vin (the gain of the
circuit) depends only on the resistors, and is
RB
RA + RB .
If RA = RB the gain is just
RB
RB
=
= 0 .5
RB + RB 2 RB
.
That is, the output voltage is exactly half the input
voltage. Expressing this gain in dB, we get:
20 log10 (0.5) = -6.02 dB.
• Notice that the answer is negative because the
output voltage is smaller than the input voltage.
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(Q6) Plot the frequency spectrum of the following
signal. In your plot remember that the ratio of
amplitudes of different frequency components is
reflected in the frequency spectrum.
Write down an equation describing the signal as a
function of time.
2
1.5
1
0.5
Voltage (V)
0
-0.5
-1
-1.5
-2
0
2
4
6
Time (ms)
8
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12
3
• Look at the signal carefully. If we draw in the dotted
line as shown….
2
1.5
1
0.5
Voltage (V)
0
-0.5
-1
-1.5
-2
0
2
4
6
Time (ms)
8
10
12
..you can see that the original signal is made up of
the sum of the following two sine waves, with peak
amplitudes 1 and 0.5 volts:
2
1.5
1
0.5
Voltage (V)
0
-0.5
-1
-1.5
-2
0
2
4
6
Time (ms)
8
10
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4
• So these two sine waves are the frequency
components of the signal.
• From the graph, we can estimate their periods (T)
– approximately 6.3 ms and 0.63 ms (notice that
ten cycles of the smaller component fit in one cycle of
the larger).
• Since f = 1/ T the frequencies are thus:
1
1
= 1580 Hz
=
158
Hz
−3
−3
and
0.63 × 10 sec
6.3 × 10 sec
• So the spectrum looks like this:
Peak amplitude
(volts)
1
0.5
0
158
Frequency (Hz)
1580
(Notice that the relative phase of the components is not
shown in the spectrum)
• The equation representing the signal will thus be
v(t) = 1 sin (2π × 158 t) + 0.5 sin (2π × 1580 t)
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For the spectrum below, sketch the individual
frequency components as a function of time, and their
sum. Use a time range of 0 – 20 ms for your plot.
You may assume that all the components are of the
form Asin(ωt) [rather than Asin(ωt + φ)]
3
Peak amplitude
(volts)
2
1
0
50
100
200
Frequency (Hz)
• Notice that the lowest frequency component is 50Hz,
corresponding to a period of 1/50 sec, or 20ms. Thus
the plot should cover one cycle of the 50 Hz
waveform.
• For the 100 Hz and 200 Hz components, there will be
2 and 4 cycles respectively.
• The peak amplitudes at 50, 100 and 200 Hz are 3.0,
1.0 and 2.0 volts respectively.
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• So the three components will look like this:
3
2
1
Voltage (V)
0
-1
-2
-3
0
0.005
0.01
0.015
0.02
0.015
0.02
Time (sec)
…and their sum will look like this:
4
2
Voltage (V) 0
-2
-4
0
0.005
0.01
Time (sec)
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