# How to measure instantaneous RF power

Arthur Pini
-June 11, 2013

Modern communications systems rely on increased signal complexity to pack more data into each symbol transmitted. These complex signals can result in data dependent, high peak-to-average power ratios requiring amplifiers and related components with large dynamic ranges. Oscilloscopes can help designers of such systems evaluate dynamic range requirements by providing a method to measure instantaneous power on very long data records. Measurement parameters can read the average and peak power levels directly from the instantaneous power waveform. The following example illustrates the techniques that can be used to measure instantaneous power.

Figure 1 shows the in-phase (I) and quadrature (Q) components of a 16-state quadrature amplitude modulated (16 QAM) baseband signal in the channel traces 1 and 2, respectively. The X-Y display shows the state transition diagram for the signal. There are distinct average power levels corresponding to the three possible combinations of the I and Q voltage levels of 0.58 and 0.18 V peak-to-peak levels.

The power levels associated with each combination of I and Q states can be determined by taking the square of the rms values of each component and adding them together. Note that since we will be taking power ratios, the power is expressed as a squared voltage,V2, to simply the measurements. If required, the square of the voltage can be divided by the impedance using the rescale function in order to read in Watts. The power levels are:

PA = 0.085 V

PB = 0.047 V

PC = 0.0083 V

The expected average power in the waveform based on the 16 states included can be calculated as:

(4 PA + 8 PB + 4 PC) /16 = 0.046 V

Figure 2 shows the steps used to measure the instantaneous power waveform and derive the peak and average power levels. The instantaneous power of the output signal is calculated by taking the sum of the squares of the I (F1) and Q (F2) waveforms. This is displayed in trace F3. The average and peak values were measured using the mean and maximum parameters reading over the sixteen states. The average power is read from the mean parameter as 46.4 mV

Based on the measured values the ratio of peak to average power is:

PPEAK/PAVG = 174/46.4 = 3.75

or 10 log (3.75) = 5.7 dB

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Figure 1 shows the in-phase (I) and quadrature (Q) components of a 16-state quadrature amplitude modulated (16 QAM) baseband signal in the channel traces 1 and 2, respectively. The X-Y display shows the state transition diagram for the signal. There are distinct average power levels corresponding to the three possible combinations of the I and Q voltage levels of 0.58 and 0.18 V peak-to-peak levels.

**Figure 1: Trace 1 shows the in-phase (I) component and Trace 2 the quadrature (Q) component of a 16 QAM signal. The state transitions are shown in the X-Y diagram. Note that the cursor marks typical peaks in the I and Q signals and simultaneously measures the phase and amplitude (radius) of the resultant vector sum**

The power levels associated with each combination of I and Q states can be determined by taking the square of the rms values of each component and adding them together. Note that since we will be taking power ratios, the power is expressed as a squared voltage,V2, to simply the measurements. If required, the square of the voltage can be divided by the impedance using the rescale function in order to read in Watts. The power levels are:

PA = 0.085 V

^{2}PB = 0.047 V

^{2}_{}PC = 0.0083 V

^{2}_{}The expected average power in the waveform based on the 16 states included can be calculated as:

(4 PA + 8 PB + 4 PC) /16 = 0.046 V

^{2}Figure 2 shows the steps used to measure the instantaneous power waveform and derive the peak and average power levels. The instantaneous power of the output signal is calculated by taking the sum of the squares of the I (F1) and Q (F2) waveforms. This is displayed in trace F3. The average and peak values were measured using the mean and maximum parameters reading over the sixteen states. The average power is read from the mean parameter as 46.4 mV

^{2}. This compares well with the expected value. The peak power is read by maximum parameter as 174 mV^{2}.

**Figure 2: The steps in computing the instantaneous power of the 16 QAM signal. The I and Q components are squared in F1 and F2. The instantaneous power is the sum of the squared values and appears in trace F3**

Based on the measured values the ratio of peak to average power is:

PPEAK/PAVG = 174/46.4 = 3.75

or 10 log (3.75) = 5.7 dB

See EDN collection: Oscilloscope articles by Arthur Pini

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