In a microwave range measurement system the electromagnetic signal v = A sin 2nft, with f = 100 MHz, is transmitted and its echo vz(1) from the target is recorded. The range is com- puted from t, the time delay between the signal and its echo. (a) Write an expression for v(1) and compute its phase angle for time delays t = 515 ns and t2 = 555 ns. (b) Can the distance be computed unambiguously from the phase angle in vz(1)? _If not, determine the additional 6.2 %3D needed information.

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6.2 In a microwave range measurement system the electromagnetic signal v1 ¼ A sin 2ft, with f ¼ 100 MHz, is transmitted and its echo v2ðtÞ from the target is recorded. The range is com- puted from , the time delay between the signal and its echo. (a) Write an expression for v2ðtÞ and compute its phase angle for time delays 1 ¼ 515 ns and 2 ¼ 555 ns. (b) Can the distance be computed unambiguously from the phase angle in v2ðtÞ? If not, determine the additional needed information. (a) Let v2ðtÞ ¼ Bsin 2f ðt  Þ ¼ B sinð2ft  - Þ. For f ¼ 100 MHz ¼ 108 Hz, - ¼ 2f  ¼ 2  108  ¼ 2k þ  where 0
In a microwave range measurement system the electromagnetic signal vị = A sin 2rft, with
f = 100 MHz, is transmitted and its echo vý(t) from the target is recorded. The range is com-
puted from t, the time delay between the signal and its echo. (a) Write an expression for v,(1)
and compute its phase angle for time delays t = 515 ns and tz = 555 ns. (b) Can the distance
be computed unambiguously from the phase angle in v,(1)? If not, determine the additional
6.2
needed information.
(a) Let vz(1) = B sin 2tf (t – t) = B sin(2rft – 6).
For f = 100 MHz = 10* Hz, 0 = 2nf t = 2 x 10°rT = 2nk +¢ where 0 < ø < 2n.
For t = 515 x 10-°, 8 = 2710* × 515 × 10-º = 1037 = 51 × 27 + ¢1 or kj = 51 and ø1 = 1.
For 1, = 555 x 10-°, 02 = 2x10* × 555 × 10-° = 1117 = 55 x 27 + ¢2 or kɔ = 55 and ø2 = .
СНAP. 6]
WAVEFORMS AND SIGNALS
117
(b) Since phase angles ø1 and ø2 are equal, the time delays 1, and t, may not be distinguished from each
other based on the corresponding phase angles ø and ø2. For unambiguous determination of the
distance, k and ø are both needed.
6.3
Show that if periods T1 and T2 of two periodic functions v1(1) and v2(1) have a common multiple,
the sum of the two functions, v(1) = v¡(t)+ v½(1), is periodic with a period equal to the smallest
common multiple of T¡ and T2. In such case show that Vavg = V1,avg +V2,avg-
%3D
If two integers n and n, can be found such that T = n¡T¡ = n,T2, then v;(1) = v;(t+n¡T¡) and
v>(t) = v2(t + n2T,). Consequently,
v(1+ T) = vj(1 + T) + vz(t + T) = v1(1) + v>(1) = v(1)
and v(1) is periodic with period T.
The average is
:| [v,(1) + v>()]dt =
+÷ v%(1) dt = V1,avg + V2.avg
V avE
Transcribed Image Text:In a microwave range measurement system the electromagnetic signal vị = A sin 2rft, with f = 100 MHz, is transmitted and its echo vý(t) from the target is recorded. The range is com- puted from t, the time delay between the signal and its echo. (a) Write an expression for v,(1) and compute its phase angle for time delays t = 515 ns and tz = 555 ns. (b) Can the distance be computed unambiguously from the phase angle in v,(1)? If not, determine the additional 6.2 needed information. (a) Let vz(1) = B sin 2tf (t – t) = B sin(2rft – 6). For f = 100 MHz = 10* Hz, 0 = 2nf t = 2 x 10°rT = 2nk +¢ where 0 < ø < 2n. For t = 515 x 10-°, 8 = 2710* × 515 × 10-º = 1037 = 51 × 27 + ¢1 or kj = 51 and ø1 = 1. For 1, = 555 x 10-°, 02 = 2x10* × 555 × 10-° = 1117 = 55 x 27 + ¢2 or kɔ = 55 and ø2 = . СНAP. 6] WAVEFORMS AND SIGNALS 117 (b) Since phase angles ø1 and ø2 are equal, the time delays 1, and t, may not be distinguished from each other based on the corresponding phase angles ø and ø2. For unambiguous determination of the distance, k and ø are both needed. 6.3 Show that if periods T1 and T2 of two periodic functions v1(1) and v2(1) have a common multiple, the sum of the two functions, v(1) = v¡(t)+ v½(1), is periodic with a period equal to the smallest common multiple of T¡ and T2. In such case show that Vavg = V1,avg +V2,avg- %3D If two integers n and n, can be found such that T = n¡T¡ = n,T2, then v;(1) = v;(t+n¡T¡) and v>(t) = v2(t + n2T,). Consequently, v(1+ T) = vj(1 + T) + vz(t + T) = v1(1) + v>(1) = v(1) and v(1) is periodic with period T. The average is :| [v,(1) + v>()]dt = +÷ v%(1) dt = V1,avg + V2.avg V avE
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