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The Image Frequency Problem in Heterodyne Receivers

By John Shin · Updated · 6 min read

The image sits 2 f_IF away from the wanted channel, on the far side of the LO, and folds onto the same IF. It must be removed before the mixer, so a higher IF eases the pre-mixer filter.

For practitionersNew to this? Plain-language version

For practitioners

Where the image comes from

A mixer multiplies the input signal by the local oscillator (LO). The product contains the sum and the difference of the two frequencies. The receiver keeps the difference and filters out the sum. The IF is therefore:

fIF=∣fin−fLO∣f_{\text{IF}} = |f_{\text{in}} - f_{\text{LO}}|

The absolute value is the cause of the problem. Two different input frequencies give the same result:

fin=fLO+fIForfin=fLO−fIFf_{\text{in}} = f_{\text{LO}} + f_{\text{IF}} \quad \text{or} \quad f_{\text{in}} = f_{\text{LO}} - f_{\text{IF}}

One of these is the wanted channel. The other is the image. The mixer does not know which one you meant. It converts both to the IF with equal enthusiasm.

Image frequency formulas

The LO can sit below or above the wanted channel.

For low-side injection, f_LO = f_RF − f_IF. The image is on the other side of the LO:

fimage=fRF−2fIFf_{\text{image}} = f_{\text{RF}} - 2 f_{\text{IF}}

For high-side injection, f_LO = f_RF + f_IF, and the image is above the wanted channel:

fimage=fRF+2fIFf_{\text{image}} = f_{\text{RF}} + 2 f_{\text{IF}}

In both cases the image and the wanted channel are exactly f_IF on either side of the LO. The spacing between them is 2 f_IF. This spacing is the quantity that decides how hard the filtering job is.

A worked example

Take f_RF = 900 MHz, f_IF = 100 MHz and low-side injection.

  • LO: f_LO = 900 − 100 = 800 MHz.
  • Image: f_image = 900 − 2 × 100 = 700 MHz.
  • Check the wanted channel: |900 − 800| = 100 MHz.
  • Check the image: |700 − 800| = 100 MHz.

Both land at 100 MHz. With high-side injection instead, the LO would be at 1000 MHz and the image at 1100 MHz. The spacing is again 200 MHz.

The figure below shows the low-side case. The two 100 MHz brackets around the LO add up to the 200 MHz bracket between image and wanted channel. The lower part of the drawing shows both inputs folding onto one IF line.

01002003004005006007008009001000Frequency (MHz)Image 700 MHzf_image = f_RF − 2 f_IFLO 800 MHzf_LO = f_RF − f_IF900 MHzwanted, f_RFf_IF100 MHzf_IF100 MHz2 f_IF = 200 MHz700 MHz (image)900 MHz (wanted)LO 800 MHzMixer100 MHzIF = |f_in − f_LO|same IF, cannot beseparated after the mixer
Image at 700 MHz and wanted channel at 900 MHz sit 2 f_IF apart around the 800 MHz LO and both fold onto the same 100 MHz IF. Concept drawing.

Why the image cannot be fixed after the mixer

Once the image is at IF, it is the same frequency as the wanted signal. Any IF filter that passes the wanted channel passes the image. Any IF filter that blocks the image blocks the wanted channel. The two have also been added together, so their separation is lost.

This leaves two options for a conventional single-mixer receiver:

  1. Remove the image before the mixer with a pre-select filter.
  2. Cancel it with an image-reject mixer architecture, which uses phase relationships to subtract the image.

The first option is the classic one. The filter sits between the antenna (or LNA) and the mixer. It must pass the wanted channel at 900 MHz and attenuate 700 MHz. The same filter also protects against other unwanted signals, but the image is usually the most demanding one because its frequency is fixed by the receiver design.

How much rejection is needed

The required image rejection follows from two things: how strong the image can be compared with the wanted signal, and the SNR the demodulator needs.

The image arrives with its own power level. After the mixer it adds to the wanted signal as interference. The pre-mixer filter must lower it until the interference is small enough for the required SNR:

Aimage≥(Pimage−Pwanted)+SNRreqA_{\text{image}} \ge \left( P_{\text{image}} - P_{\text{wanted}} \right) + \text{SNR}_{\text{req}}

Here A_image is the filter's attenuation at the image frequency in dB, minus the attenuation the filter has at the wanted frequency. Both powers are in dBm at the filter input.

As an illustration with assumed numbers, not measured ones: if an image could be 40 dB stronger than the wanted signal and the demodulator needs 10 dB SNR, the filter must give at least 50 dB more attenuation at 700 MHz than at 900 MHz. If the image can never be stronger than the wanted signal, the 10 dB requirement alone sets the rejection.

The wanted channel's own insertion loss also matters. Loss ahead of the first amplifier degrades noise figure, so a very steep filter is not free.

Choosing the IF: a trade-off

The image spacing is 2 f_IF. A larger IF puts the image farther from the wanted channel. At 900 MHz with a 100 MHz IF, the image is 200 MHz away, about 22 percent of the carrier frequency. A filter with a modest roll-off can handle that. With a much smaller IF, the image would sit close to the wanted channel, and the pre-mixer filter would need a very sharp edge.

A higher IF has a cost on the other side of the mixer. The channel-select IF filter must pick one narrow channel out of a high centre frequency. Its fractional bandwidth gets smaller, and that is harder to build.

So the IF choice splits the selectivity burden between two filters:

  • Low IF: easy channel-select filter, hard pre-mixer filter.
  • High IF: easy pre-mixer filter, hard channel-select filter.

Designers pick the IF where the two filter requirements balance, or they avoid the trade-off with an image-reject architecture.

In plain words · for students and beginners

The image problem in plain words

An everyday analogy

Imagine standing on a bridge over a river, looking at the water. A tree on the left bank and a tree on the right bank look the same in the reflection if each is the same distance from the bridge. If you only look at the reflection, you cannot say which tree is which.

The LO is the bridge. The wanted station and the mirror station are the two trees, each the same distance from it. The IF is the reflection, where both trees appear in one place.

Where the analogy breaks down: real reflections are not an exact swap. A mixer really adds both signals together at the IF, so after the mixer there is no way to pull them apart again.

The same example in simple steps

  1. The wanted station is at 900 MHz. We want the IF to be 100 MHz.
  2. We put the LO (the receiver's own tuning tone) at 800 MHz, which is 100 MHz below the station.
  3. The mixer outputs the gap between the input and the LO. For 900 MHz, the gap is 100 MHz. Good.
  4. A different station at 700 MHz is also 100 MHz away from the LO, on the other side. Its gap is also 100 MHz.
  5. So 700 MHz and 900 MHz both come out at 100 MHz. The 700 MHz one is the image.
  6. The fix is a filter before the mixer. It lets 900 MHz through and blocks 700 MHz.

The picture below puts these steps into a block diagram, from the antenna to the output.

AntennaRF pre-selectfilterMixerIF filter100 MHzOutput100 MHzLO input800 MHzremoves the image 700 MHz,passes 900 MHzBoth stations are 100 MHz from the LO, so both come out at 100 MHzImageLOWanted700 MHz800 MHz900 MHz100 MHz100 MHzSpacing is 2 f_IF = 200 MHz.Higher IF: image farther away,so the filter can be simpler.
Receiver block diagram with the pre-mixer filter stopping the 700 MHz image while passing the 900 MHz station, shown with both stations placed 100 MHz either side of the 800 MHz LO. Concept drawing.

Mistakes beginners make

  • Trying to filter the image at the IF. By then it is on the same frequency as the wanted signal. No filter can separate them.
  • Forgetting that the image moves with the LO. If you tune to a new channel, the image moves too. It is always 2 f_IF away from the wanted channel.
  • Mixing up the two injection sides. With the LO below the signal, the image is below. With the LO above, the image is above.
  • Choosing a higher IF and thinking it is free. The pre-mixer filter gets easier, but the filter that selects the channel at IF gets harder.

Remember

  • The mixer turns two input frequencies, f_LO + f_IF and f_LO − f_IF, into the same IF.
  • The image is 2 f_IF away from the wanted channel, on the far side of the LO.
  • Remove the image before the mixer. A larger IF makes that filter easier but the IF filter harder.

Frequently asked questions

What is the image frequency in a superheterodyne receiver?

It is the second input frequency that the mixer converts to the same IF as the wanted signal. It lies on the opposite side of the LO, at a distance of f_IF from the LO, so it is 2 f_IF away from the wanted channel.

How do you calculate the image frequency?

For low-side injection, f_image = f_RF − 2 f_IF. For high-side injection, f_image = f_RF + 2 f_IF. With f_RF = 900 MHz and f_IF = 100 MHz and low-side injection, the image is at 700 MHz.

Why can't the image be filtered out after the mixer?

After the mixer the wanted signal and the image are on the same IF frequency. A filter at IF cannot tell them apart, so the image must be removed before the mixer or cancelled by an image-reject architecture.

Does a higher IF help with the image problem?

Yes. The image spacing is 2 f_IF, so a higher IF moves the image farther away and the pre-mixer filter needs less selectivity. The cost is that the channel-select IF filter becomes harder to build.

References and further reading

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