The circuit shown in Figure 7.27(a) has parameters V + = 10 V , V − = − 10 V , R S = 0.5 kΩ , R E = 4 kΩ ,and R C = 2 kΩ .Thetransistor parameters are: V B E (on) = 0.7 V , V A = ∞ , and β = 100 . (a) Determine the value of C E such that the low−frequency 3 dB point is f B = 200 Hz . (b) Using the results frompart (a), determine f A . (Ans. (a) C E = 49.5 μ F , (b) f A = 0.80 Hz )
The circuit shown in Figure 7.27(a) has parameters V + = 10 V , V − = − 10 V , R S = 0.5 kΩ , R E = 4 kΩ ,and R C = 2 kΩ .Thetransistor parameters are: V B E (on) = 0.7 V , V A = ∞ , and β = 100 . (a) Determine the value of C E such that the low−frequency 3 dB point is f B = 200 Hz . (b) Using the results frompart (a), determine f A . (Ans. (a) C E = 49.5 μ F , (b) f A = 0.80 Hz )
Solution Summary: The author explains that the value of the collector current can be calculated as lI_C=V+-
The circuit shown in Figure 7.27(a) has parameters
V
+
=
10
V
,
V
−
=
−
10
V
,
R
S
=
0.5
kΩ
,
R
E
=
4
kΩ
,and
R
C
=
2
kΩ
.Thetransistor parameters are:
V
B
E
(on)
=
0.7
V
,
V
A
=
∞
, and
β
=
100
. (a) Determine the value of
C
E
such that the low−frequency 3 dB point is
f
B
=
200
Hz
. (b) Using the results frompart (a), determine
f
A
. (Ans. (a)
C
E
=
49.5
μ
F
, (b)
f
A
=
0.80
Hz
)
Can you rewrite the solution because it is
unclear?
AM
(+) = 8(1+0.5 cos 1000kt +0.5 ros 2000 thts)
=
cos 10000 πt.
8 cos wat + 4 cos wit + 4 cos Wat coswet.
J4000 t
j11000rt
$14+) = 45
jqooort
+4e
+ e
+ e
j 12000rt.
12000 kt
+ e
+e
+e
Le
jsoort
-; goon t
te
+e
Dcw>
= 885(W- 100007) + 8 IS (W-10000) -
USB
Can you rewrite the solution because it is
unclear?
Q2
AM
①(+) = 8 (1+0.5 cos 1000πt +0.5 ros 2000kt)
$4+) = 45
=
*cos 10000 πt.
8 cos wat + 4 cosat + 4 cos Wat coswet.
j1000016
+4e
-j10000πt j11000Rt
j gooort -j 9000 πt
+
e
+e
j sooort
te
+e
J11000 t
+ e
te
j 12000rt.
-J12000 kt
+ с
= 8th S(W- 100007) + 8 IS (W-10000)
<&(w) =
USB
-5-5
-4-5-4
b) Pc 2² = 64
PSB =
42
+ 4
2
Pt Pc+ PSB =
y = Pe
c) Puss =
PLSB =
= 32
4² = 8 w
32+ 8 =
× 100% = 140
(1)³×2×2
31
= 20%
x 2 = 3w
302
USB
4.5 5 5.6 6
ms Ac = 4 mi
= 0.5
mz Ac = 4
५
M2
=
=0.5
A. Draw the waveform for the following binary sequence using Bipolar RZ, Bipolar NRZ, and
Manchester code.
Data sequence= (00110100)
B. In a binary PCM system, the output signal-to-quantization ratio is to be hold to a minimum of
50 dB. If the message is a single tone with fm-5 kHz. Determine:
1) The number of required levels, and the corresponding output signal-to-quantizing noise ratio.
2) Minimum required system bandwidth.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
ENA 9.2(1)(En)(Alex) Sinusoids & Phasors - Explanation with Example 9.1 ,9.2 & PP 9.2; Author: Electrical Engineering Academy;https://www.youtube.com/watch?v=vX_LLNl-ZpU;License: Standard YouTube License, CC-BY
Electrical Engineering: Ch 10 Alternating Voltages & Phasors (8 of 82) What is a Phasor?; Author: Michel van Biezen;https://www.youtube.com/watch?v=2I1tF3ixNg0;License: Standard Youtube License