BIUST
Faculty of Engineering
Electrical, Computer and Telecommunications Department
EEN221
Tutorial for week 08
Covers material on chapter 17
Question 1
Evaluate:
(a) log10450 000
(b) log10(230)(0.9)
(c) log100.0001
Answer:
(a) log10450 000 = log104.5 × 105
= log104.5 + log10105
= 0.653 + 5.0
= 5.653
(b) log10 (230 × 0.9) = log10230 + log100.9
= 2.36 + I + 0.954
= 2.314
(c) log100.0001 = log1010−4
= −4 log1010
= −4
Question 2
Find the voltage gain in dB of a system when the applied signal is 3 mV and the output
voltage is 1.5 V.
Answer:
20 log (1.5/3mV) =
54 dB
Question 3
A low-pass RC filter circuit has a resistance of 400 Ω and a capacitor of 10 µF supplied from
a 12 V a.c. source. Find:
(a) the cut-off frequency ω2 and the magnitude of the output voltage VC at that frequency;
(b) the magnitude and phase of VC with respect to E, the source voltage, at ω = 125, 500 and
1000 rad/s.
Answer:
(a) ω2 =
1
1
=
−6
CR 10.10 ⋅ 400
= 250 rad/s
VC at 250 rad/s =
VC
=
E
1
ω 
1+  
 ω2 
2
12
2
= 8.5 V
∠ − tan −1
ω
ω2
At 125 rad/s:
(b)
VC
=
E
1
1
1+  
2
2
∠ − tan −1
1
2
= 0.894∠−26.6°
Hence VC = 10.7∠−26.6° V
At 500 rad/s:
VC
1
=
∠ − tan −1 2
2
E
1 + (2)
= 0.447∠−63.4°
Hence VC = 5.36∠−63.4° V
At 1000 rad/s:
VC
1
=
∠ − tan −1 4
2
E
1 + (4)
= 0.24∠−76°
Hence VC = 2.9∠−76° V
Question 4
A low-pass RC filter comprises a resistance of 159.2 Ω and a capacitor of 100 nF.
(a) Find the cut-off frequency f2.
(b) Draw the frequency response of the filter circuit by calculating the magnitude and phase
of the gain at frequencies of 1, 5, 10, 50 and 100 kHz. (d) cease to flow?
Answer:
f2 = 10 kHz.
At 1000 Hz:
Gain = 0.995∠−5.7°
At 5000 Hz:
Gain = 0.894∠−26.6°
At 10 000 Hz:
Gain = 0.707∠−45°
At 50 000 Hz:
Gain = 0.196∠−78.6°
At 100 000 Hz:
Gain = 0.1∠−84°
Question 5
A resistor of 10 kΩ and a capacitor of 2000 pF are used as both a low-pass and a high-pass
filter. Sketch the frequency response for both circuits. Calculate the magnitude and phase of
the gain at frequencies of half and twice the half-power frequencies for both filter
circuits.
Answer:
For both filters ωC =
1
1
=
CR 200 ⋅ 10−12 ⋅ 10 ⋅ 103
= 50 000 rad/s = 50 krad/s
For lowpass filter:
gain =
1
ω 
1+  2 
ω 
When ω =
2
∠ − tan −1
ω
ω2
ω2
2
1
gain =
1
1+  
2
2
∠ − tan −1
1
= 0.894∠−26.6°
2
When ω = 2ω2
gain =
1
1+ 2
2
∠ − tan −1 2 = 0.45∠−63.4°
For highpass filter:
gain A =
gain A =
1
 1 
1+ 

 ωCR 
1
ω 
1+  1 
ω 
2
2
∠ tan −1
∠ tan −1
ω1
ω
1
ωCR
When ω =
gain A =
ω1
2
1
1+ 2
2
∠ tan −1 2 = 0.45∠63.4°
When ω = 2ω1
1
gain A =
1
1+  
2
2
∠ tan −1
1
= 0.894∠26.6°
2
Question 6
Plot the frequency response for a high-pass RC filter circuit having a resistor of 400 Ω and a
capacitor of 0.1 µF by calculating the amplitude and phase of the gain at 2.5, 12.5, 25, 50 and
250 krad/s.
Answer:
ω1 = 25 krad/s.
At 2.5 krad/s:
Gain = 0.1∠−84°
At 12.5 krad/s:
Gain = 0.45∠−63°
At 25 krad/s:
Gain = 0.707∠−45°
At 50 krad/s:
Gain = 0.894∠−26.6°
At 250 krad/s:
Gain = 0.995∠−5.7°