A superheterodyne receiver moves the wanted signal to a fixed IF with a mixer. A mixer cannot tell which side of the LO a signal comes from, and this creates the image problem. I'll start with how the mixer produces the IF, then show where the image comes from and how the IF choice moves it. The last section shows how a receiver rejects the image. The frequency axis in all the pictures below runs from 0.0 to 3.0, and you can read the numbers as GHz.

• How does a mixer produce the IF ?
• What is the image frequency ?
• How does the IF choice move the image ?
• How is the image rejected ?

How does a mixer produce the IF ?

Before we talk about the image, we need the normal job of the mixer in view. The image is a side effect of exactly this job, so let's take one clean example first.

The picture below has two frequency axes. Axis (A) shows the mixer inputs, an LO at 1.3 and the wanted signal Sig (RF) at 1.5. Axis (B) shows the mixer output. The LO and Sig (RF) are drawn again on (B) only as a reference.

Figure 1. Mixing the signal with the LO. The mixer output has one product at the difference, the IF, and one at the sum.

• The mixer multiplies Sig (RF) at 1.5 with the LO at 1.3.
• The product labelled Sig (IF) sits at 1.5 - 1.3 = 0.2. This is the IF frequency, and it equals the distance between the LO and Sig (RF), which the IF arrow on axis (B) marks.
• The second Mixer Out product sits at 1.5 + 1.3 = 2.8. The IF filter after the mixer removes it.

So the IF follows f_IF = |f_RF - f_LO|, and the sum product sits at f_RF + f_LO. Both terms come from the product of two cosines, because cos(a)cos(b) equals (cos(a - b) + cos(a + b))/2. In Figure 1 the LO is below the signal, which is called low-side injection. With the LO above the signal, at 1.7, the IF would also be 0.2. That is high-side injection.

• A mixer outputs the difference and the sum : with the LO at 1.3 and the signal at 1.5, the products are 0.2 and 2.8.
• The IF is the distance between the LO and the signal : f_IF = |f_RF - f_LO|.
• The LO can sit below or above the signal : both low-side and high-side injection give the same IF.

What is the image frequency ?

The absolute value in the IF formula causes the problem. Two different input frequencies give the same distance to the LO, one on each side. So let's see what happens to a signal on the other side.

The picture below repeats the setup of Figure 1, with the same LO at 1.3 and Sig (RF) at 1.5. Axis (C) now has one more input, a black block labelled Image at 1.1. Axis (D) shows what the mixer makes of the image.

Figure 2. The image frequency. A signal at 2 x IF from the wanted signal, on the other side of the LO, lands on exactly the same IF.

• The Image at 1.1 is 0.2 below the LO, and Sig (RF) is 0.2 above it. Both have the same distance to the LO.
• On axis (D) the image produces Image (IF) at 1.3 - 1.1 = 0.2. It sits right on top of Sig (IF), so no filter after the mixer can separate the two.
• The image also produces a sum product at 1.1 + 1.3 = 2.4, which the IF filter removes.
• The arrows under axis (D) mark IF on each side of the LO, and 2 x IF from the image to Sig (RF).

So for low-side injection the image sits at f_image = f_LO - f_IF = f_RF - 2f_IF. For high-side injection it sits at f_RF + 2f_IF. In both cases the image is 2 x IF away from the wanted signal. Any energy at that frequency, such as another operator's carrier or noise, lands on the wanted IF. It also adds the noise of the image band to the IF, so an unfiltered image raises the noise figure of the receiver. For a noiseless mixer with equal gain for both bands, the image band doubles the noise at the IF output, which is 3 dB more noise.

• The image is the mirror of the signal around the LO : with LO 1.3 and signal 1.5, the image is at 1.1.
• The image and the signal are 2 x IF apart : f_image = f_RF - 2f_IF for low-side injection.
• After the mixer the image cannot be removed : it lands on the same IF as the wanted signal.
• The image band also brings noise : without rejection, the noise of the image band adds to the IF.

How does the IF choice move the image ?

The image is always 2 x IF away from the signal, so the IF is the parameter that sets how hard the image is to remove. The next two pictures raise the IF from 0.2 to 0.4 and show what moves.

The picture below keeps Sig (RF) at 1.5 but moves the LO down to 1.1. Axis (E) shows the inputs, and axis (F) shows the mixer output.

Figure 3. A higher IF. With the LO at 1.1, the IF becomes 0.4.

• Sig (IF) now sits at 1.5 - 1.1 = 0.4, and the sum product at 1.5 + 1.1 = 2.6.
• The IF arrow between the LO and Sig (RF) is twice as long as in Figure 1.

The picture below adds the image for this new LO. Axis (G) has a black block at 0.7, which is 0.4 below the LO at 1.1. Axis (H) shows where it lands.

Figure 4. The image for a higher IF. The image moves to 0.7, which is 0.8 away from the wanted signal instead of 0.4.

• The black block at 0.7 mixes with the LO at 1.1 and lands at 1.1 - 0.7 = 0.4, on top of Sig (IF).
• Its sum product sits at 0.7 + 1.1 = 1.8, marked Mixer Out on axis (H).
• Compared with Figure 2, the image is now twice as far from Sig (RF).

Let's compare Figure 2 and Figure 4. The image is still there, and it still lands on the IF. But with the higher IF it is 0.8 away from the signal instead of 0.4, so a filter in front of the mixer has more room to attenuate it. This is the classic trade-off of a superheterodyne receiver. A high IF makes the image easy to reject in front of the mixer. A low IF makes the channel filter at the IF easier to build, because a narrow filter is easier at a low centre frequency. Some receivers use two conversions to get both.

• The image distance grows with the IF : an IF of 0.2 puts the image 0.4 away, and an IF of 0.4 puts it 0.8 away.
• A higher IF eases the image filter : the filter in front of the mixer gets a wider transition band.
• A lower IF eases the channel filter : a narrow filter is easier to build at a low centre frequency.
• Double conversion combines both : a high first IF for the image, and a low second IF for the channel selection.

How is the image rejected ?

The image must be removed before the mixer, because after the mixer it overlaps the wanted IF. So the rejection happens at RF, either with a filter or with a mixer structure that cancels the image.

The picture below goes back to the setup of Figure 2, with the LO at 1.3, Sig (RF) at 1.5 and the Image at 1.1. Axis (I) shows the inputs. Axis (J) shows an Image Rejection Filter, whose passband covers Sig (RF) and whose lower skirt falls over the image. Axis (K) shows the input after the filter, with a much smaller Filtered Image. Axis (L) shows the mixer output.

Figure 5. Image rejection with a filter. The filter attenuates the image before the mixer, so only a small image product lands on the IF.

• The Image Rejection Filter on axis (J) passes the band around Sig (RF) and falls off below it, where the image sits.
• On axis (K) the Filtered Image is much smaller than the Image on axis (I). Sig (RF) and the LO keep their size.
• On axis (L), Image (IF) at 0.2 is now a small block under Sig (IF). Its sum product at 2.4 is small too.
• The marks IF, IF and 2 x IF under axis (L) show that the frequencies are the same as in Figure 2. Only the image level has changed.

How much rejection is enough? Let's take an example. Assume the signal in the image band is 30 dB stronger than the wanted signal, and the demodulator needs 10 dB of SNR. Then the image product must end up at least 10 dB below the wanted IF. So the filter has to attenuate the image by 30 + 10 = 40 dB or more. With an IF of 0.2, the filter must fall by that amount between 1.5 and 1.1, and a steeper skirt means a larger and more lossy filter.

The other method is an image reject mixer. It uses an I mixer and a Q mixer with LOs 90 deg apart, and it combines the two outputs with another 90 deg shift. In this combination the wanted signal adds and the image cancels. The cancellation depends on how well the two paths match. When the gain ratio of the two paths is g and the phase error is φ, the image rejection ratio is (1 + 2g cosφ + g²) / (1 - 2g cosφ + g²). A gain error of 1 dB and a phase error of 5 deg give about 22.8 dB. A gain error of 0.1 dB and a phase error of 1 deg give about 39.6 dB. So a practical receiver often combines an image reject mixer with a filter. The same I/Q mismatch appears in a direct conversion receiver, as the Homodyne/Zero IF/Direct Conversion page explains.

• The image is removed at RF, before the mixer : after the mixer it overlaps the wanted IF.
• The required rejection adds the image level and the required SNR : an image 30 dB above the signal with a 10 dB SNR target needs 40 dB.
• An image reject mixer cancels the image with I and Q paths : its rejection depends on the gain and phase match of the two paths.
• 1 dB and 5 deg of mismatch give only about 22.8 dB : 0.1 dB and 1 deg give about 39.6 dB.