1.1 Suppose the input to an amplifier is xa (t) = sin(2πF0 t) and the steady-state output is
ya (t) = 100 sin(2πF0 t + φ1 ) − 2 sin(4πF0 t + φ2 ) + cos(6πF0 t + φ3 )
(a) Is the amplifier a linear system or is it a nonlinear system?
(b) What is the gain of the amplifier?
(c) Find the average power of the output signal.
(d) What is the total harmonic distortion of the amplifier?
Solution
(a) The amplifier is nonlinear because the steady-state output contains harmonics.
(b) From (1.1.2), the amplifier gain is K = 100.
(c) From (1.2.4), the output power is
d20 1 2
d1 + d+ 22 + d23
Py = +
4 2
= .5(1002 + 22 + 1)
= 5002.5
(d) From (1.2.5)
100(Py − d21 /2)
THD =
Py
100(5002.5 − 5000)
=
5002.5
= .05%
√
1.2 Consider the following signum function that returns the sign of its argument.
1 , t>0
∆
sgn(t) = 0 , t=0
−1 , t < 0
1
c 2017 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, (a) Using Appendix 1, find the magnitude spectrum.
(b) Find the phase spectrum.
Solution
(a) From Table A2 in Appendix 1
1
Xa (f ) =
jπf
Thus the magnitude spectrum is
Aa (f ) = |Xa(f )|
1
=
|jπf |
1
=
π|f |
(b) The phase spectrum is
φa (f ) = 6 Xa (f )
= −6 jπf
π
= −sgn(f )
2
1.3 Parseval’s identity states that a signal and its spectrum are related in the following way.
Z ∞ Z ∞
|xa (t)|2 dt = |Xa(f )|2 df
−∞ −∞
Use Parseval’s identity to compute the following integral.
Z ∞
J = sinc2 (2Bt)dt
−∞
2
c 2017 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, Solution
From Table A2 in Appendix 1 if
xa (t) = sinc(2Bt)
then
µa (f + B) − µa (f − B)
Xa (f ) =
2B
Thus by Parseval’s identity
Z ∞
J = sin2 (2Bt)dt
Z−∞
∞
= |xa (t)|2 dt
Z−∞
∞
= |Xa(f )|2 df
−∞
B
1
Z
= df
2B −B
= 1
1.4 Consider the causal exponential signal
xa (t) = exp(−ct)µa (t)
(a) Using Appendix 1, find the magnitude spectrum.
(b) Find the phase spectrum
(c) Sketch the magnitude and phase spectra when c = 1.
3
c 2017 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, Solution
(a) From Table A2 in Appendix 1
1
Xa (f ) =
c + j2πf
Thus the magnitude spectrum is
Aa(f ) = |Xa(f )|
1
=
|c + j2πf |
1
= p
c + (2πf )2
2
(b) The phase spectrum is
Aa (f ) = |Xa (f )|
= 1 − 6 (c + j2πf )
6
−1 2πf
= − tan
c
4
c 2017 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.