In the figure below, the spectrum analyzer displays the frequency spectrum of a

Question

In the figure below, the spectrum analyzer displays the frequency spectrum of a periodic square-wave signal. The spectrum analyzer range (span) is set from O Hz to 15 kHz. What is the relationship between the periodicity (T) of the periodic square-wave and the locations of the spectral lines peaks of the frequency spectrum? For example, the first spectral line labeled in the figure as the "fundamental" (first harmonic) has the following relationship 1/T (1 over the the periodicity or period time), the next spectral line is at /T, the next spectral line is at /T, the next spectral line is at /T, the next spectral line is at /T,the next spectral line is at /T and the next spectral line is at /T. The general relationship is /T. Hint: odd or even integer. Oscilloscope-XSC1 23 1 , | In Use List … xSc1Oscilloscope Spectrum Analyzer-XSA1 Span control Zero span Full span Enter kHz Hz kHz Range: 1 Span: 15 Start: 0 Center: 7.5 End: 15 Time -+| 27.000 ms T2 +11 27.000 ms 5.000 V 5.000 V Ref: 0 dB fundamental T2-T1 0.000s 0.000 Resolution freq: 58.594 Hz Timebase Scale: 500 us/Div X pos. (Div): 0 Channel A 58.594 Hz Scale: 2 Y pos.Div): Istarti Reverse show refer,! set. 8.009 kHz 11.936 m nput Trigger

Explanation / Answer

Answer:- Relationship-

The spectral peaks of sqaure wave occurs at odd multiples of 1/T .

So first spectral line is at 1/T, second is at 3/T, 3rd spetral line is at 5/T, next will be at 7/T and so on.

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