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As discussed in class, when a rectangular window is used on a signal observed for 7 seconds, the signal's spectrum is distorted due to a convolution with a sinc function whose main lobe has width 4. This convolution causes smearing of the signal's spectrum, which limits the resolution with which we can observe it. rad/sec, to be the frequency resolution of Let us consider the half-width of this main lobe, spectral analysis using a rectangular window.¹ (a) Suppose a signal is observed for T = 10 seconds. What is the achievable frequency resolution (in Hz) if we use a rectangular window? (b) Suppose a signal is observed for T = 100 seconds. What is the achievable frequency resolution (in Hz) if we use a rectangular window? (c) How long would we need to observe a signal with sinusoidal components at 138.100 Hz and 138.101 Hz so that their difference in frequency (0.001 Hz) is equal to the achievable frequency resolution using a rectangular window? (d) Let T denote your answer to part (c). How many cycles will a sinusoid of frequency 138.100 Hz complete in T seconds? How many cycles will a sinusoid of frequency 138.101 Hz complete in T seconds? How different are your answers? We see that T seconds is barely long enough for these two sinusoids to distinguish themselves.

As discussed in class, when a rectangular window is used on a signal observed for 7 seconds, the signal's spectrum is distorted due to a convolution with a sinc function whose main lobe has width 4. This convolution causes smearing of the signal's spectrum, which limits the resolution with which we can observe it. rad/sec, to be the frequency resolution of Let us consider the half-width of this main lobe, spectral analysis using a rectangular window.¹ (a) Suppose a signal is observed for T = 10 seconds. What is the achievable frequency resolution (in Hz) if we use a rectangular window? (b) Suppose a signal is observed for T = 100 seconds. What is the achievable frequency resolution (in Hz) if we use a rectangular window? (c) How long would we need to observe a signal with sinusoidal components at 138.100 Hz and 138.101 Hz so that their difference in frequency (0.001 Hz) is equal to the achievable frequency resolution using a rectangular window? (d) Let T denote your answer to part (c). How many cycles will a sinusoid of frequency 138.100 Hz complete in T seconds? How many cycles will a sinusoid of frequency 138.101 Hz complete in T seconds? How different are your answers? We see that T seconds is barely long enough for these two sinusoids to distinguish themselves.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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As discussed in class, when a rectangular window is used on a signal observed for T seconds, the signal's spectrum is distorted due to a convolution with a sinc function whose main lobe has width. This convolution causes smearing of the signal's spectrum, which limits the resolution with which we can observe it. Let us consider the half-width of this main lobe, 2 rad/sec, to be the frequency resolution of spectral analysis using a rectangular window.¹ (a) Suppose a signal is observed for T = 10 seconds. What is the achievable frequency resolution (in Hz) if we use a rectangular window? (b) Suppose a signal is observed for T = 100 seconds. What is the achievable frequency resolution (in Hz) if we use a rectangular window? (c) How long would we need to observe a signal with sinusoidal components at 138.100 Hz and 138.101 Hz so that their difference in frequency (0.001 Hz) is equal to the achievable frequency resolution using a rectangular window? (d) Let T denote your answer to part (c). How many cycles will a sinusoid of frequency 138.100 Hz complete in T seconds? How many cycles will a sinusoid of frequency 138.101 Hz complete in T seconds? How different are your answers? We see that T seconds is barely long enough for these two sinusoids to distinguish themselves.
Transcribed Image Text:As discussed in class, when a rectangular window is used on a signal observed for T seconds, the signal's spectrum is distorted due to a convolution with a sinc function whose main lobe has width. This convolution causes smearing of the signal's spectrum, which limits the resolution with which we can observe it. Let us consider the half-width of this main lobe, 2 rad/sec, to be the frequency resolution of spectral analysis using a rectangular window.¹ (a) Suppose a signal is observed for T = 10 seconds. What is the achievable frequency resolution (in Hz) if we use a rectangular window? (b) Suppose a signal is observed for T = 100 seconds. What is the achievable frequency resolution (in Hz) if we use a rectangular window? (c) How long would we need to observe a signal with sinusoidal components at 138.100 Hz and 138.101 Hz so that their difference in frequency (0.001 Hz) is equal to the achievable frequency resolution using a rectangular window? (d) Let T denote your answer to part (c). How many cycles will a sinusoid of frequency 138.100 Hz complete in T seconds? How many cycles will a sinusoid of frequency 138.101 Hz complete in T seconds? How different are your answers? We see that T seconds is barely long enough for these two sinusoids to distinguish themselves.
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