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2 2 2 Let v t Vdc Vac cos !t  . Show that Veff Vdc 1 Vac . 2
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2 Veff
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1 T V Vac cos !t  2 dt T 0 dc 1 T 2 2 V Vac cos2 !t  2Vdc Vac cos !t  dt T 0 dc
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2 2 Vdc 1 Vac 2
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2 Veff hv2 t i h Vdc Vac cos !t  2 i 2 2 hVdc Vac cos2 !t  2Vdc Vac cos !t  i 2 2 Vdc Vac hcos2 !t  i 2Vdc Vac hcos !t  i 2 2 Vdc 1 Vac 2
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Let f1 and f2 be two di erent harmonicsq of f0 . Show that the e ective value of 2 2 v t V1 cos 2f1 t 1 V2 cos 2f2 t 2 is 1 V1 V2 . 2
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2 2 v2 t V1 cos2 2f1 t 1 V2 cos2 2f2 t 2
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2V1 V2 cos 2f1 t 1 cos 2f2 t 2
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2 2 2 Veff hv2 t i V1 hcos2 2f1 t 1 i V2 hcos2 2f2 t 2 i
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2V1 V2 hcos 2f1 t 1 cos 2f2 t 2 i But hcos 2f1 t 1 i hcos 2f2 t 2 i 1=2 (see Problem 6.4) and
WAVEFORMS AND SIGNALS
[CHAP. 6
hcos 2f1 t 1 cos 2f2 t 2 i
1 hcos 2 f1 f2 t 1 2 i 2 1 hcos 2 f1 f2 t 1 2 i 0 2
2 2 2 Therefore, Veff 1 V1 V2 and Veff 2
q 1 2 2 2 V1 V2 :
The signal v t in Fig. 6-16 is sinusoidal. Find its period and frequency. v t A B cos !t  and nd its average and rms values.
Express it in the form
Fig. 6-16 The time between two positive peaks, T 20 s, is one period corresponding to a frequency f 0:05 Hz. The signal is a cosine function with amplitude B added to a constant value A. B 1 Vmax Vmin 1 8 4 6 2 2 A Vmax B Vmin B 2
The cosine is shifted by 2 s to the right, which corresponds to a phase lag of 2=20 3608 368. Therefore, the signal is expressed by   t 368 v t 2 6 cos 10 The average and e ective values are found from A and B: Vavg A 2;
2 Veff A2 B2 =2 22 62 =2 22
Veff
p 22 4:69
Let v1 cos 200t and v2 cos 202t. Show that v v1 v2 is periodic. Find its period, Vmax , and the times when v attains its maximum value.
The periods of v1 and v2 are T1 1=100 s and T2 1=101 s, respectively. The period of v v1 v2 is the smallest common multiple of T1 and T2 , which is T 100T1 101T2 1 s. The maximum of v occurs at t k with k an integer when v1 and v2 are at their maxima and Vmax 2.
Convert v t 3 cos 100t 4 sin 100t to A sin 100t  .
p p Note that 3= 32 42 3=5 sin 36:878 and 4= 32 42 4=5 cos 36:878.
Then,
v t 3 cos 100t 4 sin 100t 5 0:6 cos 100t 0:8 sin 100t 5 sin 36:878 cos 100t cos 36:878 sin 100t 5 sin 100t 36:878
CHAP. 6]
WAVEFORMS AND SIGNALS
Find the average and e ective value of v2 t in Fig. 6-1(b) for V1 2, V2 1, T 4T1 .
V2;avg
2 V2;eff
V1 T1 V2 T T1 V1 3V2 0:25 4 T 2 2 p V T V2 T T1 7 1 1 or V2;eff 7=2 1:32 T 4
Find V3;avg and V3;eff in Fig. 6-1(c) for T 100T1 .
2 From Fig. 6-1(c), V3;avg 0. To nd V3;eff , observe that the integral of v2 over one period is V0 T1 =2. 3 2 The average of v3 over T 100T1 is therefore p 2 2 2 or V3;eff V0 2=20 0:0707V0 hv2 t i V3;eff V0 T1 =200T1 V0 =200 3 p The e ective value of the tone burst is reduced by the factor T=T1 10.
Referring to Fig. 6-1(d), let T 6 and let the areas under the positive and negative sections of v4 t be 5 and 3, respectively. Find the average and e ective values of v4 t .
V4;avg 5 3 =6 1=3 The e ective value cannot be determined from the given data.
Find the average and e ective value of the half-recti ed cosine wave v1 t shown in Fig. 6-17(a).
V1;avg
2 V1;eff
Vm T
2 Vm T 2 Vm 2T
from which V1;eff Vm =2.
  2t V T 2t T=4 V m dt m sin T 2T T T=4  T=4   T=4 2t V 2 T=4 4t dt m dt cos2 1 cos T T 2T T=4 T=4     T 4t T=4 V2 T T V2 sin m t m 4 T T=4 2T 4 4 4 T=4 cos
Find the average and e ective value of the full-recti ed cosine wave v2 t Vm j cos 2t=Tj shown in Fig. 6-17(b).
Fig. 6-17 Use the results of Problems 6.3 and 6.13 to nd V2;avg . v2 t v1 t v1 t T=2 and Thus,
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