(a) Consider the circuit shown in Figure 12.3 ( a ) . Assume A 1 = 100 and A 2 = 10. Determine the output signal-to-noise ratio in terms of the input signal-to-noise ratio. (b) Consider the circuit shown in Figure 12.3 ( c ) . Assume A 1 = 10 4 , A 2 = 10 , and β = 0.001. Determine the output signal-to-noise ratio in terms of the input signal-to-noise ratio. (Ans. (a) S o / N o = 100 ( S i / N i ) , (b) S o / N o = 10 4 ( S i / N i ) )
(a) Consider the circuit shown in Figure 12.3 ( a ) . Assume A 1 = 100 and A 2 = 10. Determine the output signal-to-noise ratio in terms of the input signal-to-noise ratio. (b) Consider the circuit shown in Figure 12.3 ( c ) . Assume A 1 = 10 4 , A 2 = 10 , and β = 0.001. Determine the output signal-to-noise ratio in terms of the input signal-to-noise ratio. (Ans. (a) S o / N o = 100 ( S i / N i ) , (b) S o / N o = 10 4 ( S i / N i ) )
Solution Summary: The author calculates the output signal to noise ratio in terms of the input signal-to-noise ratio. The first amplifier is fed with an input v_i which gets amplified by the first
(a) Consider the circuit shown in Figure
12.3
(
a )
. Assume
A
1
=
100
and
A
2
=
10.
Determine the output signal-to-noise ratio in terms of the input signal-to-noise ratio. (b) Consider the circuit shown in Figure
12.3
(
c
)
.
Assume
A
1
=
10
4
,
A
2
=
10
,
and
β
=
0.001.
Determine the output signal-to-noise ratio in terms of the input signal-to-noise ratio. (Ans. (a)
S
o
/
N
o
=
100
(
S
i
/
N
i
)
, (b)
S
o
/
N
o
=
10
4
(
S
i
/
N
i
)
)
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.
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