Page 128 - Schaum's Outline of Theory and Problems of Electric Circuits

WAVEFORMS AND SIGNALS
117
CHAP. 6]
(b) Since phase angles 1 and 2 are equal, the time delays 1 and 2 may not be distinguished from each
other based on the corresponding phase angles 1 and 2 . For unambiguous determination of the
distance, k and are both needed.

6.3 Show that if periods T 1 and T 2 of two periodic functions v 1 ðtÞ and v 2 ðtÞ have a common multiple,
the sum of the two functions, vðtÞ¼ v 1 ðtÞþ v 2 ðtÞ, is periodic with a period equal to the smallest
common multiple of T 1 and T 2 . In such case show that V avg ¼ V 1;avg þ V 2;avg .
If two integers n 1 and n 2 can be found such that T ¼ n 1 T 1 ¼ n 2 T 2 , then v 1 ðtÞ¼ v 1 ðt þ n 1 T 1 Þ and
v 2 ðtÞ¼ v 2 ðt þ n 2 T 2 Þ. Consequently,
vðt þ TÞ¼ v 1 ðt þ TÞþ v 2 ðt þ TÞ¼ v 1 ðtÞþ v 2 ðtÞ¼ vðtÞ

and vðtÞ is periodic with period T.
The average is
ð ð ð
1 T 1 T 1 T
V avg ¼ ½v 1 ðtÞþ v 2 ðtފ dt ¼ v 1 ðtÞ dt þ v 2 ðtÞ dt ¼ V 1;avg þ V 2;avg
T 0 T 0 T 0

2
6.4 Show that the average of cos ð!t þ Þ is 1/2.
1
2
Using the identity cos ð!t þ Þ¼ ½1 þ cos 2ð!t þ ފ, the notation h f i¼ F avg , and the result of
2
Problem 6.3, we have
h1 þ cos 2ð!t þ Þi ¼ h1iþhcos 2ð!t þ Þi
2
But hcos 2ð!t þ Þi ¼ 0. Therefore, hcos ð!t þ Þi ¼ 1=2.

2 2 1 2
6.5 Let vðtÞ¼ V dc þ V ac cos ð!t þ Þ. Show that V eff ¼ V þ V ac .
dc
2
ð T
1
2 2
V eff ¼ ½V dc þ V ac cos ð!t þ ފ dt
T 0
ð
1 T
2
2
2
¼ ½V dc þ V ac cos ð!t þ Þþ 2V dc V ac cos ð!t þ ފ dt
T 0
2 1 2
¼ V dc þ V ac
2
Alternatively, we can write
2 2 2
V eff ¼hv ðtÞi ¼ h½V dc þ V ac cos ð!t þ ފ i
2 2 2
¼hV dc þ V ac cos ð!t þ Þþ 2V dc V ac cos ð!t þ Þi
2 2 2
¼ V dc þ V ac hcos ð!t þ Þi þ 2V dc V ac hcos ð!t þ Þi
2 1 2
¼ V dc þ V ac
2
6.6 Let f 1 and f 2 be two different harmonics of f 0 . Show that the effective value of
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
vðtÞ¼ V 1 cosð2 f 1 t þ 1 Þþ V 2 cos ð2 f 2 t þ 2 Þ is 1 ðV þ V Þ.
2 1 2
2
2
2
2
2
v ðtÞ¼ V 1 cos ð2 f 1 t þ 1 Þþ V 2 cos ð2 f 2 t þ 2 Þ
þ 2V 1 V 2 cos ð2 f 1 t þ 1 Þ cos ð2 f 2 t þ 2 Þ
2 2 2 2 2 2
V eff ¼hv ðtÞi ¼ V 1 hcos ð2 f 1 t þ 1 Þi þ V 2 hcos ð2 f 2 t þ 2 Þi
þ 2V 1 V 2 hcos ð2 f 1 t þ 1 Þ cos ð2 f 2 t þ 2 Þi
2
2
But hcos ð2 f 1 t þ 1 Þi ¼ hcos ð2 f 2 t þ 2 Þi ¼ 1=2 (see Problem 6.4) and