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Homework Help: Frequency Spectrum of a Real Sine Wave

  1. Feb 13, 2010 #1
    Hi. I'm doing a lab where we hooked up an RF signal generator at 1.25MHz/10dBm to a spectrum analyzer using a 50 ohm wire.

    Can anyone explain to me or link me to a place where I can read about why a real sine wave's frequency spectrum is not a pulse? Also, why would with a sweep from 1 Mhz to 4 Mhz the SA show three sine waves separated by multiples of the frequency of the intended sine wave(~-10dBw @ 1.25, ~-66 dBw @ 2.5, ~-87 dBw @ 3.75)? Would there also be lowered amplitude frequencies to the left of the intended frequency?

    I'm pretty confused, because I haven't taken the signals and systems class that stresses frequency response, FFT, and so on. Also, I have no idea how this hardware works. I just want an informed lab report!
  2. jcsd
  3. Feb 13, 2010 #2


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    Because you didn't measure a sine wave and cannot do so, not even in principle. Measuring a sine wave would require doing the measurement *forever* i.e. infinitely into the past and into the future. This is another way of saying that an actual sine wave is infinite in the time domain and hence no real signal will be a sine wave. If you measure a sinusoidal signal over some finite time interval, what you are essentially measuring is a sine wave multiplied by a so called "top hat" or "rectangular" function (a step function that is equal to 0 everywhere outside the time interval measured and equal to 1 inside that time interval). The abrupt change at either end of the time interval introduces (a whole continuum of) higher frequency components. Another way of looking at it: this signal is clearly aperiodic, hence it doesn't have a single frequency. When you do your signals and systems course, you'll learn more about how to figure out what the Fourier transform of the resulting signal would look like. If you're curious, look up the 'multiplication property' of the Fourier transform. The punchline is that a multiplication of two signals in the time domain, when Fourier transformed, will result in a convolution of the individual Fourier transforms of those two signals in the frequency domain. What shape or profile did the output of the spectrum analyzer have?

    I'm not sure why you're seeing those higher frequency harmonics. I wonder if it's some sampling or aliasing issue.
  4. Feb 18, 2010 #3
    The actual signal from the sig gen isn't perfectly narrow in frequency - the reality is that all sig gens have rounded response rather than a perfect spike.

    Even an expensive sig gen isn't spectrally pure - there are always some sidebands or harmonics which is what you are seeing. If you dig up the sig gen's data sheet, you'll likely see specs defining how big these sidebands can be and likely (if calibrated and working properly) the values you are seeing are below those spec's.

    If the spectrum analyzer can't measure down to those specs you might not see that however. Assuming otherwise above.

    There are also artifacts that can be caused by the way swept-tuned spectrum analyzers or FFT-based analyzers work that spread the otherwise impulse-like frequency spurs to look more rounded. This is usually caused by sweeping too fast relative to the resolution bandwidth - generally using the automatic modes avoid this. Don't go to manual settings on the spectrum analyzer without knowing how. Agilent has a Fundamentals of Spectrum Analysis application note that goes into every you need to know about this. Highly recommended.
  5. Feb 18, 2010 #4
    Thanks, guys. I think I turned in a decent report. I talked mainly about the SG's inability to produce perfect sine waves(evidenced by the harmonics spotted) and about how the filter resolution also probably widened the pulse.
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