RANDOM SIGNALS

CHAPTER 17 Fourier Method of Waveform Analysis 420

6.12 RANDOM SIGNALS

So far we have dealt with signals which are completely specified. For example, the values of a sinusoidal waveform, such as the line voltage, can be determined for all times if its amplitude, frequency, and phase are known. Such signals are calleddeterministic.

There exists another class of signals which can be specified only partly through their time averages, such as their mean, rms value, and frequency range. These are calledrandom signals. Random signals can carry information and should not be mistaken with noise, which normally corrupts the information contents of the signal.

The voltage recorded at the terminals of a microphone due to speech utterance and the signals picked up by an antenna tuned to a radio or TV station are examples of random signals. The future course and values of such signals can be predicted only in average and not precisely. Other examples of random signals are the binary waveforms in digital computers, image intensities over the area of a picture, and the speech or music which modulates the amplitude of carrier waves in an AM system.

It may not seem useful to discuss signals whose values are specified only in average. However, through harmonic analysis we can still find much about the average effect of such signals in electric circuits.

EXAMPLE 6.25 Samples from a random signalxðtÞare recorded every 1 ms and designated byxðnÞ. Approx- imate the mean and rms values ofxðtÞfrom samples given in Table 6-2.

Fig. 6-15

The time averages ofxðtÞandx²ðtÞmay be approximated fromxðnÞ.

X_avg¼ ð2þ4þ11þ5þ7þ6þ9þ10þ3þ6þ8þ4þ1þ3þ5þ12Þ=16¼6

X_eff² ¼ ð2²þ4²þ11²þ5²þ7²þ6²þ9²þ10²þ3³þ6²þ8²þ4²þ1²þ3²þ5²þ12²Þ=16¼46 X_eff ¼6:78

EXAMPLE 6.26 A binary signalvðtÞis either at 0.5 or0:5 V. It can change its sign at 1-ms intervals. The sign change is not known a priori, but it has an equal chance for positive or negative values. Therefore, if measured for a long time, it spends an equal amount of time at the 0.5-V and0:5-V levels. Determine its average and effective values over a period of 10 s.

During the 10-s period, there are 10,000 intervals, each of 1-ms duration, which on average are equally divided between the 0.5-V and0:5-V levels. Therefore, the average ofvðtÞcan be approximated as

v_avg¼ ð0:550000:55000Þ=10,000¼0 The effective value ofvðtÞis

V_eff² ¼ ½ð0:5Þ²5000þ ð0:5Þ²5000=10,000¼ ð0:5Þ² or V_eff¼0:5 V The value ofV_eff is exact and independent of the number of intervals.

Solved Problems

6.1 Find the maximum and minimum values ofv¼1þ2 sinð!tþÞ, given!¼1000 rad/s and¼3 rad. Determine if the functionvis periodic, and find its frequencyf and periodT. Specify the phase angle in degrees.

V_max¼1þ2¼3 V_min¼12¼ 1

The functionv is periodic. To find the frequency and period, we note that!¼2f ¼1000 rad/s.

Thus,

f ¼1000=2¼159:15 Hz and T¼1=f ¼2=1000¼0:00628 s¼6:28 ms Phase angle¼3 rad¼18083=¼171:98

6.2 In a microwave range measurement system the electromagnetic signal v₁ ¼Asin 2ft, with f ¼100 MHz, is transmitted and its echov₂ðtÞ from the target is recorded. The range is com- puted from, the time delay between the signal and its echo. (a) Write an expression forv₂ðtÞ and compute its phase angle for time delays₁¼515 ns and₂¼555 ns. (b) Can the distance be computed unambiguously from the phase angle in v₂ðtÞ? If not, determine the additional needed information.

(a) Letv₂ðtÞ ¼Bsin 2fðtÞ ¼Bsinð2ftÞ.

Forf ¼100 MHz¼10⁸Hz,¼2f¼210⁸¼2kþwhere 0< <2.

For₁¼51510⁹,₁¼210⁸51510⁹¼103¼512þ₁ork₁¼51 and₁¼. For₂¼55510⁹,₂¼210⁸55510⁹¼111¼552þ₂ork₂¼55 and₂¼.

Table 6-2

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

xðnÞ 2 4 11 5 7 6 9 10 3 6 8 4 1 3 5 12

(b) Since phase angles₁and₂are equal, the time delays₁and₂may not be distinguished from each other based on the corresponding phase angles₁ and₂. For unambiguous determination of the distance,kandare both needed.

6.3 Show that if periodsT₁andT₂of two periodic functionsv₁ðtÞandv₂ðtÞhave a common multiple, the sum of the two functions,vðtÞ ¼v₁ðtÞ þv₂ðtÞ, is periodic with a period equal to the smallest common multiple ofT₁ andT₂. In such case show thatV_avg¼V_1;avgþV_2;avg.

If two integers n₁ and n₂ can be found such that T¼n₁T₁¼n₂T₂, then v₁ðtÞ ¼v₁ðtþn₁T₁Þ and v₂ðtÞ ¼v₂ðtþn₂T₂Þ. Consequently,

vðtþTÞ ¼v₁ðtþTÞ þv₂ðtþTÞ ¼v₁ðtÞ þv₂ðtÞ ¼vðtÞ andvðtÞis periodic with periodT.

The average is V_avg¼ 1

T ðT

0

½v₁ðtÞ þv₂ðtÞdt¼ 1 T

ðT 0

v₁ðtÞdtþ1 T

ðT 0

v₂ðtÞdt¼V_1;avgþV_2;avg

6.4 Show that the average of cos²ð!tþÞis 1/2.

Using the identity cos²ð!tþÞ ¼¹₂½1þcos 2ð!tþÞ, the notation hfi ¼F_avg, and the result of Problem 6.3, we have

h1þcos 2ð!tþÞi ¼ h1i þ hcos 2ð!tþÞi Buthcos 2ð!tþÞi ¼0. Therefore,hcos²ð!tþÞi ¼1=2.

6.5 LetvðtÞ ¼V_dcþV_accosð!tþÞ. Show thatV_eff² ¼V_dc² þ¹₂V_ac². V_eff² ¼ 1

T ðT

0

½V_dcþV_accosð!tþÞ²dt

¼ 1 T

ðT 0

½V_dc² þV_ac² cos²ð!tþÞ þ2V_dcV_accosð!tþÞdt

¼V_dc² þ¹₂V_ac² Alternatively, we can write

V_eff² ¼ hv²ðtÞi ¼ h½V_dcþV_accosð!tþÞ²i

¼ hV_dc² þV_ac² cos²ð!tþÞ þ2V_dcV_accosð!tþÞi

¼V_dc² þV_ac²hcos²ð!tþÞi þ2V_dcV_achcosð!tþÞi

¼V_dc² þ¹₂V_ac²

6.6 Let f₁ and f₂ be two different harmonics of f₀. Show that the effective value of vðtÞ ¼V₁cosð2f₁tþ₁Þ þV₂cosð2f₂tþ₂Þis

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1

2ðV₁²þV₂²Þ q

. v²ðtÞ ¼V₁²cos²ð2f₁tþ₁Þ þV₂²cos²ð2f₂tþ₂Þ

þ2V₁V₂cosð2f₁tþ₁Þcosð2f₂tþ₂Þ

V_eff² ¼ hv²ðtÞi ¼V₁²hcos²ð2f₁tþ₁Þi þV₂²hcos²ð2f₂tþ₂Þi þ2V₁V₂hcosð2f₁tþ₁Þcosð2f₂tþ₂Þi

Buthcos²ð2f₁tþ₁Þi ¼ hcos²ð2f₂tþ₂Þi ¼1=2 (see Problem 6.4) and

hcosð2f₁tþ₁Þcosð2f₂tþ₂Þi ¼ 1

2hcos½2ðf₁þf₂Þtþ ð₁þ₂Þi þ1

2hcos½2ðf₁f₂Þtþ ð₁₂Þi ¼0 Therefore,V_eff² ¼¹₂ðV₁²þV₂²ÞandV_eff ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1

2ðV₁²þV₂²Þ q

:

6.7 The signalvðtÞin Fig. 6-16 is sinusoidal. Find its period and frequency. Express it in the form vðtÞ ¼AþBcosð!tþÞand find its average and rms values.

The time between two positive peaks,T¼20 s, is one period corresponding to a frequencyf ¼0:05 Hz.

The signal is a cosine function with amplitudeBadded to a constant valueA.

B¼¹₂ðV_maxV_minÞ ¼¹₂ð8þ4Þ ¼6 A¼V_maxB¼V_minþB¼2

The cosine is shifted by 2 s to the right, which corresponds to a phase lag ofð2=20Þ3608¼368. Therefore, the signal is expressed by

vðtÞ ¼2þ6 cos 10t368

The average and effective values are found fromAandB:

V_avg¼A¼2; V_eff² ¼A²þB²=2¼2²þ6²=2¼22 or V_eff ¼ ffiffiffiffiffi

22 p

¼4:69

6.8 Letv₁¼cos 200tandv₂¼cos 202t. Show thatv¼v₁þv₂is periodic. Find its period,V_max, and the times whenv attains its maximum value.

The periods ofv₁andv₂areT₁¼1=100 s andT₂¼1=101 s, respectively. The period ofv¼v₁þv₂is the smallest common multiple ofT₁andT₂, which isT¼100T₁¼101T₂¼1 s. The maximum ofvoccurs att¼kwithkan integer whenv₁andv₂are at their maxima andV_max¼2.

6.9 ConvertvðtÞ ¼3 cos 100tþ4 sin 100ttoAsinð100tþÞ.

Note that 3= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3²þ4²

p ¼3=5¼sin 36:878and 4= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3²þ4²

p ¼4=5¼cos 36:878. Then, vðtÞ ¼3 cos 100tþ4 sin 100t¼5ð0:6 cos 100tþ0:8 sin 100tÞ

¼5ðsin 36:878cos 100tþcos 36:878sin 100tÞ ¼5 sinð100tþ36:878Þ Fig. 6-16

6.10 Find the average and effective value ofv₂ðtÞin Fig. 6-1(b) forV₁¼2,V₂¼1,T ¼4T₁. V_2;avg¼V₁T₁V₂ðTT₁Þ

T ¼V₁3V₂

4 ¼ 0:25 V_2;eff² ¼V₁²T₁þV₂²ðTT₁Þ

T ¼7

4 or V_2;eff¼ ffiffiffi

7 p

=2¼1:32

6.11 FindV_3;avgandV_3;eff in Fig. 6-1(c) forT¼100T₁.

From Fig. 6-1(c),V_3;avg¼0. To findV_3;eff, observe that the integral ofv²₃over one period isV₀²T₁=2.

The average ofv²₃overT¼100T₁is therefore

hv²₃ðtÞ i ¼V_3;eff² ¼V₀²T₁=200T₁¼V₀²=200 or V_3;eff ¼V₀ ffiffiffi 2 p

=20¼0:0707V₀ The effective value of the tone burst is reduced by the factor ffiffiffiffiffiffiffiffiffiffiffiffi

T=T₁

p ¼10.

6.12 Referring to Fig. 6-1(d), letT ¼6 and let the areas under the positive and negative sections of v₄ðtÞbeþ5 and 3, respectively. Find the average and effective values ofv₄ðtÞ.

V_4;avg¼ ð53Þ=6¼1=3 The effective value cannot be determined from the given data.

6.13 Find the average and effective value of the half-rectified cosine wavev₁ðtÞshown in Fig. 6-17(a).

V_1;avg¼V_m T

ðT=4 T=4

cos2t

T dt¼V_mT

2T sin2t T

T=4

T=4

¼V_m V_1;eff² ¼V_m²

T ðT=4

T=4

cos²2t

T dt¼V_m² 2T

ðT=4 T=4

1þcos4t T

dt

¼V_m² 2T tþ T

4sin4t T

T=4

T=4

¼V_m² 2T

T 4þT

4

¼V_m² 4 from whichV_1;eff ¼V_m=2.

6.14 Find the average and effective value of the full-rectified cosine wavev₂ðtÞ ¼V_mjcos 2t=Tjshown in Fig. 6-17(b).

Use the results of Problems 6.3 and 6.13 to findV_2;avg. Thus,

v₂ðtÞ ¼v₁ðtÞ þv₁ðtT=2Þ and V_2;avg¼V_1;avgþV_1;avg¼2V_1;avg¼2V_m= Use the results of Problems 6.5 and 6.13 to findV_2;eff. And so,

V_2;eff² ¼V_1;eff² þV_1;eff² ¼2V_1;eff² ¼V_m²=2 or V_2;eff ¼V_m= ffiffiffi 2 p Fig. 6-17

The rms value ofv₂ðtÞcan also be derived directly. Because of the squaring operation, a full-rectified cosine function has the same rms value as the cosine function itself, which isV_m= ffiffiffi

2 p

.

6.15 A 100-mH inductor in series with 20- resistor [Fig. 6-18(a)] carries a current i as shown in Fig. 6-18(b). Find and plot the voltages acrossR,L, andRL.

i¼ 10

10ð110³tÞ ðAÞ 0

and di

dt¼

0 fort<0

10⁴A=s for 0<t<10³s 0 fort>10³s 8>

<

>: 8>

<

>:

v_R¼Ri¼ 200 V

200ð110³tÞ ðVÞ 0

and v_L¼Ldi dt¼

0 fort<0 1000 V for 0<t<10³s 0 fort>10³s 8>

<

>: 8>

<

>:

Since the passive elements are in series,v_RL¼v_Rþv_L and so

v_RL¼

200 V fort<0

2ð10⁵tÞ 800 ðVÞ for 0<t<10³s

0 fort>10³s

8<

:

The graphs of v_L andv_RL are given in Fig. 6-18(c) and (d), respectively. The plot of the resistor voltagev_Rhas the same shape as that of the current [see Fig. 6-18(b)], except for scaling by a factor ofþ20.

Fig. 6-18

6.16 A radar signal sðtÞ, with amplitude V_m¼100 V, consists of repeated tone bursts. Each tone burst lastsT_b¼50ms. The bursts are repeated every T_s¼10 ms. Find S_eff and the average power insðtÞ.

LetV_eff ¼V_m ffiffiffi p2

be the effective value of the sinusoid within a burst. The energy contained in a single burst isW_b¼T_bV_eff². The energy contained in one period ofsðtÞisW_s¼T_sS_eff² . SinceW_b¼W_s¼W, we obtain

T_bV_eff² ¼T_sS²_eff S_eff² ¼ ðT_b=T_sÞV_eff² S_eff ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi T_b=T_s

p V_eff ð40Þ

Substituting the values ofT_b,T_s, andV_eff into (40), we obtain S_eff ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð5010⁶Þ=ð1010³Þ q

ð100= ffiffiffi 2 p

Þ ¼5 V ThenW¼10²ð25Þ ¼0:25 J. The average power insðtÞis

P¼W=T_s¼T_sS²_eff=T_s¼S²_eff¼25 W

The average power ofsðtÞ is represented byS²_eff and its peak power byV_eff² . The ratio of peak power to average power is ffiffiffiffiffiffiffiffiffiffiffiffiffi

T_s=T_b

p . In this example the average power and the peak power are 25 W and 5000 W, respectively.

6.17 An appliance usesV_eff ¼120 V at 60 Hz and drawsI_eff¼10 A with a phase lag of 608. Express v,i, andp¼vias functions of time and show that power is periodic with a dc value. Find the frequency, and the average, maximum, and minimum values ofp.

v¼120 ffiffiffi 2 p

cos!t i¼10 ffiffiffi 2 p

cosð!t608Þ

p¼vi¼2400 cos!tcosð!t608Þ ¼1200 cos 608þ1200 cosð2!t608Þ ¼600þ1200 cosð2!t608Þ The power function is periodic. The frequency f ¼260¼120 Hz and P_avg¼600 W, p_max¼600þ 1200¼1800 W,p_min¼6001200¼ 600 W.

6.18 A narrow pulsei_sof 1-A amplitude and 1-ms duration enters a 1-mF capacitor att¼0, as shown in Fig. 6-19. The capacitor is initially uncharged. Find the voltage across the capacitor.

The voltage across the capacitor is

V_C¼ 1 C

ðt 1

i dt¼

0 fort<0

10⁶t ðVÞ for 0<t<1ms (charging period) 1 V fort>1ms

8<

:

If the same amount of charge were deposited on the capacitor in zero time, then we would havev¼uðtÞ (V) andiðtÞ ¼10⁶ðtÞ(A).

6.19 The narrow pulse i_s of Problem 6.18 enters a parallel combination of a 1-mF capacitor and a 1-M resistor (Fig. 6-20). Assume the pulse ends at t¼0 and that the capacitor is initially uncharged. Find the voltage across the parallelRC combination.

Fig. 6-19

Letvdesignate the voltage across the parallelRCcombination. The current inRisi_R¼v=R¼10⁶v.

During the pulse,i_Rremains negligible becausevcannot exceed 1 V andi_Rremains under 1mA. Therefore, it is reasonable to assume that during the pulse,i_C¼1 A and consequentlyvð0^þÞ ¼1 V. Fort>0, from application of KVL around theRCloop we get

vþdv

dt¼0; vð0^þÞ ¼1 V ð41Þ

The only solution to (41) isv¼e^tfort>0 orvðtÞ ¼e^tuðtÞfor allt. For all practical purposes,i_scan be considered an impulse of size 10⁶A, and thenv¼e^tuðtÞ(V) is called theresponseof theRCcombination to the current impulse.

6.20 Plot the functionvðtÞwhich varies exponentially from 5 V att¼0 to 12 V att¼ 1with a time constant of 2 s. Write the equation forvðtÞ.

Identify the initial pointA(t¼0 andv¼5Þand the asymptotev¼12 in Fig. 6-21. The tangent atA intersects the asymptote att¼2, which is pointBon the line. Draw the tangent lineAB. Identify pointC belonging to the curve att¼2. For a more accurate plot, identify pointDatt¼4. Draw the curve as shown. The equation isvðtÞ ¼Ae^t=2þB. From the initial and final conditions, we getvð0Þ ¼AþB¼5 andvð1Þ ¼B¼12 orA¼ 7, andvðtÞ ¼ 7e^t=2þ12.

6.21 The voltagev¼V₀e^ajtjfora>0 is connected across a parallel combination of a resistor and a capacitor as shown in Fig. 6-22(a). (a) Find currentsi_C,i_R, andi¼i_Cþi_R. (b) Compute and graphv,i_C,i_R, andiforV₀¼10 V,C¼1mF,R¼1 M, anda¼1.

(a) See (a) in Table 6-3 for the required currents.

(b) See (b) in Table 6-3. Figures 6-22(b)–(e) show the plots ofv,i_C,i_R, andi, respectively, for the given data. Duringt>0,i¼0, and the voltage source does not supply any current to theRCcombination.

The resistor current needed to sustain the exponential voltage across it is supplied by the capacitor.

Fig. 6-20

Fig. 6-21

Fig. 6-22

Supplementary Problems

6.22 Let v₁¼8 sin 100t and v₂¼6 sin 99t. Show that v¼v₁þv₂ is periodic. Find the period, and the maximum, average, and effective values ofv. Ans: T¼2;V_max¼14;V_avg¼0;V_eff ¼5 ffiffiffi

2 p

6.23 Find period, frequency, phase angle in degrees, and maximum, minimum, average, and effective values of vðtÞ ¼2þ6 cosð10tþ=6Þ.

Ans: T¼0:2 s;f ¼5 Hz;phase¼308;V_max¼8;V_min¼ 4;V_avg¼2;V_eff ¼ ffiffiffiffiffi p22

6.24 ReducevðtÞ ¼2 cosð!tþ308Þ þ3 cos!ttovðtÞ ¼Asinð!tþÞ. Ans: A¼4:84; ¼1028 6.25 FindV_2;avg andV_2;eff in the graph of Fig. 6-1(b) forV₁¼V₂¼3, andT¼4T₁=3.

Ans: V_2;avg¼1:5;V_2;eff ¼3

6.26 Repeat Problem 6.25 forV₁¼0,V₂¼4, andT¼2T₁. Ans: V_2;avg¼ 2;V_2;eff¼2 ffiffiffi 2 p

6.27 FindV_3;avg andV_3;eff in the graph of Fig. 6-1(c) forV₀¼2 andT¼200T₁. Ans: V_3;avg¼0;V_3;eff¼0:1

6.28 The waveform in Fig. 6-23 is sinusoidal. Express it in the formv¼AþBsinð!tþÞand find its mean and rms values. Ans: vðtÞ ¼1þ6 sinðt=12þ1208Þ;V_avg¼1;V_eff ¼ ffiffiffiffiffi

19 p Table 6-3

Time v i_C¼C dv=dt i_R¼v=R i¼i_Cþi_R (a)

t<0 t>0

v¼V₀e^at v¼V₀e^at

i_C¼CV₀ae^at i_C¼ CV₀ae^at

i_R¼ ðV₀=RÞe^at i_R¼ ðV₀=RÞe^at

i¼v₀ðCaþ1=RÞe^at i¼V₀ðCaþ1=RÞe^at (b)

t<0 t>0

v¼10e^t v¼10e^t

i_C¼10⁵e^t i_C¼ 10⁵e^t

i_R¼10⁵e^t i_R¼10⁵e^t

i¼2ð10⁵e^tÞ i¼0

Fig. 6-23

6.29 Find the average and effective values ofv₁ðtÞin Fig. 6-24(a) andv₂ðtÞin Fig. 6-24(b).

Ans: V_1;avg¼ 1

3;V_1;eff ¼ ffiffiffiffiffi 17 3 r

; V_2;avg¼ 1

2;V_2;eff¼ ffiffiffiffiffi 13 2 r

6.30 The current through a seriesRLcircuit withR¼5andL¼10 H is given in Fig. 6-10(a) whereT¼1 s.

Find the voltage acrossRL.

Ans: v¼

0 fort<0 10þ5t for 0<t<1 5 fort>1 8>

<

>:

6.31 Find the capacitor current in Problem 6.19 (Fig. 6-20) for allt. Ans: i_C¼10⁶½ðtÞ e^tuðtÞ

6.32 The voltagevacross a 1-H inductor consists of one cycle of a sinusoidal waveform as shown in Fig. 6-25(a).

(a) Write the equation forvðtÞ. (b) Find and plot the current through the inductor. (c) Find the amount and time of maximum energy in the inductor.

Ans: ðaÞ v¼ ½uðtÞ uðtTÞsin2t T ðVÞ ðbÞ i¼ ðT=2Þ½uðtÞ uðtTÞ 1cos2t

T

ðAÞ: See Fig. 6-25ðbÞ:

ðcÞ W_max¼ 1

2²T² ðJÞ att¼T=2

6.33 Write the expression forvðtÞwhich decays exponentially from 7 att¼0 to 3 att¼ 1with a time constant of 200 ms. Ans: vðtÞ ¼3þ4e^5tfort>0

6.34 Write the expression forvðtÞwhich grows exponentially with a time constant of 0.8 s from zero att¼ 1to 9 att¼0. Ans: vðtÞ ¼9e^5t=4fort<0

6.35 Express the current of Fig. 6-6 in terms of step functions.

Ans: iðtÞ ¼4uðtÞ þ6X¹

k¼1

½uðt5kÞ uðt5kþ2Þ

6.36 In Fig. 6-10(a) letT¼1 s and call the waveforms₁ðtÞ. Expresss₁ðtÞand its first two derivativesds₁=dtand d²s₁=dt², using step and impulse functions.

Ans: s₁ðtÞ ¼ ½uðtÞ uðt1Þtþuðt1Þ;ds₁=dt¼uðtÞ uðt1Þ;d²s₁=dt²¼ðtÞ ðt1Þ Fig. 6-24

6.37 Find an impulse voltage which creates a 1-A current jump att¼0 when applied across a 10-mH inductor.

Ans: vðtÞ ¼10²ðtÞ(V)

6.38 (a) Given v₁¼cost, v₂¼cosðtþ308Þ andv¼v₁þv₂, write v in the form of a single cosine function v¼AcosðtþÞ. (b) Find effective values ofv₁,v₂, andv. Discuss whyV_eff² >ðV_1;eff² þV_2;eff² Þ.

Ans. (a) v¼1:93 cosðtþ158Þ; ðbÞ V_1;eff¼V_2;eff ¼0:707;V_eff ¼1:366V_eff is found from the following derivation

V_eff² ¼ hv²i ¼ hðv₁þv₂Þ²i ¼ hv²₁þv²₂þ2v₁v₂i ¼ hv²₁i þ hv²₂i þ2hv₁v₂i

Sincev₁andv₂ have the same frequency and are 308out of phase, we gethV₁V₂i ¼¹₂cos 308¼ ffiffiffi p3

=4, which is positive. Therefore,V_eff² >ðV_1;eff² þV_2;eff² Þ:

6.39 (a) Show that v₁¼costþcos ffiffiffi 2 p

t is not periodic. (b) Replace ffiffiffi 2 p

by 1.4 and then show that v₂¼costþcos 1:4tis periodic and find its periodT₂. (c) Replace ffiffiffi

p2

by 1.41 and find the periodT₃of v₃¼costþcos 1:41t. (d) Replace ffiffiffi

2 p

by 1.4142 and find the periodT₄ofv₄¼costþcos 1:4142t.

Ans. (a) ffiffiffi p2

is not a rational number. Therefore,v₁is not periodic. (b) T₂¼10s. (c) T₃¼200s.

(d) T₄¼10 000s.

6.40 A random signalsðtÞwith an rms value of 5 V has a dc value of 2 V. Find the rms value ofs₀ðtÞ ¼sðtÞ 2, that is, when the dc component is removed. Ans: S_0;eff ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi

5²4 p

¼ ffiffiffiffiffi p21

¼4:58 V Fig. 6-25

127

First-Order Circuits