Solution of ECE 202 Test 11 S13
1.
A resonant circuit has a Q of 40 and a center frequency of 5 kHz. What is its numerical
bandwidth in Hz?
Bandwidth = ____________ Hz
B=
2.
f0 5000
=
= 125 Hz
Q
40
A lowpass filter has an input impedance of 300 ohms and a corner frequency of 35 kHz. It is
desired to change its input impedance to 50 ohms and its corner frequency to 1.4 MHz.
(a)
By what numerical factor must each resistor value be multiplied?
Resistance factor = ____________
50 1
= = 0.16667
300 6
(b)
By what numerical factor must each inductor value be multiplied?
Inductance factor = ____________
50
35, 000
1
×
=
= 0.0042
300 1, 400, 000 240
(c)
By what numerical factor must each capacitor value be multiplied?
Capacitance factor = ____________
300
35, 000
3
×
=
= 0.15
50 1, 400, 000 20
3.
The output voltage in a circuit is 30 dB above the input voltage. If the input voltage is 20 mV
what is the numerical output voltage in volts?
Output Voltage = ____________
Vo
⎛ Vo ⎞
20 log10 ⎜
= 30 ⇒
= 10 30 /20 ⇒ Vo = 632.5mV
⎝ 20mV ⎟⎠
20mV
4.
A circuit transfer function H ( s ) has one pole at s = −300 and no finite zeros. If its magnitude at
1 rad/s is 35 dB, what is its magnitude in dB at 3 × 10 6 rad/s?
(
H j3 × 10 6
)
= ____________ dB
The corner frequency is ω = 300 . One rad/s is very far below the corner so the response is
practically flat at 35 dB there. 3 × 10 6 rad/s is well above the corner so the response at 3 × 10 6
rad/s is practically the same as the Bode diagram asymptote there. 3 × 10 6 rad/s is 4 decades above
300 rad/s. Therefore the response at 3 × 10 6 rad/s is down by 80 dB from 35 dB which is -45 dB.
Solution of ECE 202 Test 11 S13
1.
A resonant circuit has a Q of 60 and a center frequency of 3 kHz. What is its numerical
bandwidth in Hz?
Bandwidth = ____________ Hz
B=
2.
f0 3000
=
= 50 Hz
Q
60
A lowpass filter has an input impedance of 400 ohms and a corner frequency of 50 kHz. It is
desired to change its input impedance to 50 ohms and its corner frequency to 6 MHz.
(a)
By what numerical factor must each resistor value be multiplied?
Resistance factor = ____________
50 1
= = 0.125
400 8
(b)
By what numerical factor must each inductor value be multiplied?
Inductance factor = ____________
50
50, 000
1
1
1
×
= ×
=
= 0.0010416
400 6, 000, 000 8 120 960
(c)
By what numerical factor must each capacitor value be multiplied?
Capacitance factor = ____________
400
50, 000
1
1
×
=8×
=
= 0.0667
50 6, 000, 000
120 15
3.
The output voltage in a circuit is 25 dB above the input voltage. If the input voltage is 10 mV
what is the numerical output voltage in volts?
Output Voltage = ____________
Vo
⎛ V ⎞
20 log10 ⎜ o ⎟ = 25 ⇒
= 10 25 /20 ⇒ Vo = 177.83mV
⎝ 10mV ⎠
10mV
4.
A circuit transfer function H ( s ) has one pole at s = −500 and no finite zeros. If its magnitude at
1 rad/s is 45 dB, what is its magnitude in dB at 5 × 10 5 rad/s?
(
H j5 × 10 5
)
= ____________ dB
The corner frequency is ω = 500 . One rad/s is very far below the corner so the response is
practically flat at 45 dB there. 5 × 10 5 rad/s is well above the corner so the response at 5 × 10 5
rad/s is practically the same as the Bode diagram asymptote there. 5 × 10 5 rad/s is 3 decades above
500 rad/s. Therefore the response at 5 × 10 5 rad/s is down by 60 dB from 45 dB which is -15 dB.
Solution of ECE 202 Test 11 S13
1.
A resonant circuit has a Q of 50 and a center frequency of 12 kHz. What is its numerical
bandwidth in Hz?
Bandwidth = ____________ Hz
B=
2.
f0 12000
=
= 240 Hz
Q
50
A lowpass filter has an input impedance of 200 ohms and a corner frequency of 15 kHz. It is
desired to change its input impedance to 50 ohms and its corner frequency to 4 MHz.
(a)
By what numerical factor must each resistor value be multiplied?
Resistance factor = ____________
50 1
= = 0.25
200 4
(b)
By what numerical factor must each inductor value be multiplied?
Inductance factor = ____________
50
15, 000
×
= 0.0009375
200 4, 000, 000
(c)
By what numerical factor must each capacitor value be multiplied?
Capacitance factor = ____________
200
15, 000
×
= 0.015
50 4, 000, 000
3.
The output voltage in a circuit is 15 dB above the input voltage. If the input voltage is 50 mV
what is the numerical output voltage in volts?
Output Voltage = ____________
Vo
⎛ Vo ⎞
20 log10 ⎜
= 15 ⇒
= 1015 /20 ⇒ Vo = 281.2mV
⎝ 50mV ⎟⎠
50mV
4.
A circuit transfer function H ( s ) has one pole at s = −200 and no finite zeros. If its magnitude at
1 rad/s is 16 dB, what is its magnitude in dB at 2 × 10 6 rad/s?
(
H j2 × 10 6
)
= ____________ dB
The corner frequency is ω = 200 . One rad/s is very far below the corner so the response is
practically flat at 16 dB there. 2 × 10 6 rad/s is well above the corner so the response at 2 × 10 6
rad/s is practically the same as the Bode diagram asymptote there. 2 × 10 6 rad/s is 4 decades above
200 rad/s. Therefore the response at 2 × 10 6 rad/s is down by 80 dB from 16 dB which is -64 dB.