Most of the available spectrum snalysers do not want a DC component to their 50 ohm input. But how low should the cutoff frequency be? Also which parameters indicates that clearly in the datasheets? As this clearly depends on the applied amplitude.

Lets say for an example a first order high pass filter with fc= 1 Hz and Rinput=50 Ohm: C=1/(2pi*50 Ohm * 1 Hz) = 3183 uF

  • \$\begingroup\$ Ok I maybe found something: DC coupled: 10 Hz to 4 GHz; AC coupled: 1 MHz to 4 GHz. But then again, how should the attenuation be, below that 10 Hz? \$\endgroup\$ – Jeres Jun 26 at 6:35
  • \$\begingroup\$ why would it matter? it's AC-coupled, i.e. there's a built-in high-pass filter. So, any (not dangerously high) DC offset doesn't matter. \$\endgroup\$ – Marcus Müller Jun 26 at 7:51
  • \$\begingroup\$ Marcus: for the simple reason that I want to measure down to 10 Hz, where it is DC coupled.. And it says MAX 0V DC. \$\endgroup\$ – Jeres Jun 26 at 8:05
  • \$\begingroup\$ well, then you already have an answer, don't you? Your filter needs to have a cutoff below 10 Hz, that's all. But, honestly, using a spectrum analyzer to measure something at 10 Hz sounds like a bad idea and you should probably ask a new question titled "how do I measure {this specific property} of a low-frequency {description of signal}", where you describe what the goal of your measurement is, what you want to figure out based on that measurement. \$\endgroup\$ – Marcus Müller Jun 26 at 8:17
  • \$\begingroup\$ Also, note how your first order input filter doesn't work: it doesn't consitute a wideband 50Ω match (I'm assuming that if you're using a spectrum analyzer, you care about multiple decades of bandwidth). \$\endgroup\$ – Marcus Müller Jun 26 at 8:23

There are two reasons that DC and a spectrum analyser generally don't mix.

The first 'easy to understand' reason is that their 50 Ω input impedance will be a real 50 Ω resistor, or an attenuator with real resistors, and too much power will damage it. You don't need many volts DC input for 100 mW or so, only a little over 2 V. Many signals that you'll want to measure will be riding on a higher voltage than that.

If you want to measure below 1 MHz, then you'll need to use the DC coupled input, with an external high pass filter. The filter is only stopping DC from overheating the input resistor, so you can choose its passband based on what you want to measure.

The second 'only obvious once you've used a spectrum analyser at very low frequencies' reason is that they're very bad near DC.

The balance of the first mixer is all that stands in the way of the first local oscillator (LO) leakage appearing also at DC, and the first LO close to carrier phase noise appearing at low frequencies. These signals are often very large indeed, often higher than the top of the screen, unless your particular analyser does an active balance of the mixer, in which case they can be got onscreen for a while until the temperature drifts. This is why many analysers don't let you scan down to DC, or don't specify DC, even on a DC coupled input.

Your analyser specifies only 10 Hz and above when DC coupled. If you need to measure below 10 Hz, you'll need a different instrument. If you choose a 1 Hz corner frequency, that will work, and you'll get good measurements above 10 Hz. But between 1 Hz and 10 Hz? You might be able to infer something from what the analyser does down there, but it's not specified to produce believable results in that range.

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  • \$\begingroup\$ Thanks Neil! I want to measure the noise floor(10 Hz...500 kHz) of an Opamp. That is why I choose the Signal Analyser which can measure down to 10 Hz, which is sufficient. So fc=5Hz will be enough? \$\endgroup\$ – Jeres Jun 26 at 9:58

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